101. At fixed above-threshold frequency, increasing the intensity of radiation on a metal surface mainly increases
ⓐ. the number of photoelectrons emitted per second
ⓑ. the work function of the metal
ⓒ. the maximum kinetic energy of each photoelectron
ⓓ. the threshold frequency of the metal
Correct Answer: the number of photoelectrons emitted per second
Explanation: At fixed frequency above the threshold frequency, each photon has the same energy \(h\nu\). Increasing intensity means more energy is incident per unit area per unit time. Since the photon energy is unchanged, higher intensity corresponds to a larger number of photons arriving per second. More suitable photons can eject more electrons per second, so the photoelectric current increases. The maximum kinetic energy does not increase because the energy of each photon is not changed.
102. Two current-potential curves are obtained for the same metal and same incident frequency above threshold.
Curve \(P\) and curve \(Q\) cut the negative potential axis at the same point. Curve \(P\) reaches a higher saturation current than curve \(Q\).
The best interpretation is that curve \(P\) corresponds to
ⓐ. lower frequency and lower intensity
ⓑ. higher intensity but the same photon energy
ⓒ. higher work function of the same metal
ⓓ. lower threshold frequency caused by illumination
Correct Answer: higher intensity but the same photon energy
Explanation: The same negative-axis intercept means the stopping potential is the same for both curves. Since stopping potential depends on \(K_{\max}\), the maximum kinetic energy of photoelectrons is unchanged. For the same metal, this indicates that the photon energy and frequency are the same. The higher saturation current of curve \(P\) means more photoelectrons are emitted and collected per second. At fixed frequency above threshold, that increase is explained by higher intensity.
103. For the same metal and same above-threshold frequency, the intensity is changed from \(I\) to \(4I\). If the initial saturation current is \(3.0\,\mu A\), the new saturation current is expected to be
ⓐ. \(0.75\,\mu A\)
ⓑ. \(3.0\,\mu A\)
ⓒ. \(12.0\,\mu A\)
ⓓ. \(6.0\,\mu A\)
Correct Answer: \(12.0\,\mu A\)
Explanation: \( \textbf{Given:} \) Initial intensity \(I\), final intensity \(4I\), and initial saturation current \(I_s=3.0\,\mu A\).
\( \textbf{Condition:} \) Frequency is fixed and above \( \nu_0 \).
\( \textbf{Useful proportionality:} \)
\[
I_s\propto \text{intensity}
\]
\( \textbf{Reason:} \) At fixed frequency, photon energy remains constant, so intensity change changes mainly the photon rate.
\( \textbf{Current ratio:} \)
\[
\frac{I_{s2}}{I_{s1}}=\frac{4I}{I}=4
\]
\( \textbf{Substitution:} \)
\[
I_{s2}=4(3.0\,\mu A)
\]
\( \textbf{Calculation:} \)
\[
I_{s2}=12.0\,\mu A
\]
\( \textbf{Final answer:} \) The new saturation current is \(12.0\,\mu A\), while the stopping potential would remain unchanged.
104. A statement says, “Doubling the intensity of above-threshold light doubles the stopping potential.” The most accurate correction is that doubling intensity
ⓐ. doubles stopping potential but leaves current unchanged
ⓑ. makes the threshold frequency equal to zero
ⓒ. reduces photon energy to half
ⓓ. raises saturation current but leaves \(V_0\) unchanged
Correct Answer: raises saturation current but leaves \(V_0\) unchanged
Explanation: For fixed frequency above threshold, each photon carries the same energy \(h\nu\). Doubling intensity increases the number of photons incident per second, so more photoelectrons may be emitted per second. This increases the saturation current. The stopping potential depends on the maximum kinetic energy of photoelectrons, which depends on photon energy and work function. Since the frequency is unchanged, the stopping potential remains unchanged even though more electrons are emitted.
105. Consider the following statements for fixed incident frequency \( \nu\gt\nu_0 \).
Statement I: Photoelectric current increases when intensity increases.
Statement II: Saturation current is proportional to intensity under the same frequency condition.
Statement III: Stopping potential increases in the same ratio as intensity.
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I and II only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Statement I is true because higher intensity at fixed suitable frequency increases the number of photons arriving per second. This can increase the number of photoelectrons emitted per second and hence the photoelectric current. Statement II is also true because saturation current represents the maximum collected electron rate, which follows the emission rate. Statement III is false because stopping potential depends on \(K_{\max}\), not on the number of electrons emitted. Intensity changes the quantity of emitted electrons, not the maximum energy per emitted electron.
106. Below-threshold radiation is directed at a metal surface. If only the intensity is made extremely large, the expected photoelectric current is
ⓐ. a large saturation current
ⓑ. equal to the stopping potential
ⓒ. zero
ⓓ. independent of the metal used
Correct Answer: zero
Explanation: If \( \nu\lt\nu_0 \), each photon has energy less than the work function of the metal. Increasing intensity increases the number of photons, but it does not increase the energy of each photon. In the basic photoelectric effect, an electron absorbs energy from a single photon for emission. Since each photon is still insufficient, photoelectrons are not emitted. The threshold condition cannot be overcome by intensity alone when the frequency remains below threshold.
107. A data record from a photoelectric experiment is shown below.
| Trial | Frequency | Intensity | Saturation current | Stopping potential |
|---|
| P | \(\nu\gt\nu_0\) | \(I\) | \(i_s\) | \(V_0\) |
| Q | \(\nu\gt\nu_0\) | \(3I\) | \(3i_s\) | \(V_0\) |
| R | \(\nu\lt\nu_0\) | \(5I\) | \(0\) | No stopping potential measured |
The record mainly shows that
ⓐ. intensity affects current only above threshold
ⓑ. intensity controls emission even below threshold
ⓒ. stopping potential is proportional to intensity
ⓓ. threshold frequency depends on saturation current
Correct Answer: intensity affects current only above threshold
Explanation: Trials \(P\) and \(Q\) have the same above-threshold frequency, so increasing intensity increases the saturation current. Their stopping potential remains the same because the photon energy is unchanged. Trial \(R\) has frequency below threshold, so even high intensity does not produce photoemission. This shows that intensity can increase the number of emitted electrons only when individual photons are energetic enough. The frequency condition is therefore the first requirement for photoelectric emission.
108. Assertion: At fixed frequency above threshold, increasing intensity increases photoelectric current.
Reason: Increasing intensity at fixed frequency increases the number of incident photons per second.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The Assertion is true when the incident frequency is above the threshold frequency. At fixed frequency, photon energy \(h\nu\) remains constant. Increasing intensity means a larger photon rate reaches the metal surface. A larger number of suitable photons can eject a larger number of electrons per second. The Reason correctly explains why the current increases without implying any increase in maximum kinetic energy.
109. Radiation of frequency \( \nu_1 \) just above \( \nu_0 \) is replaced by radiation of higher frequency \( \nu_2 \), while the same metal is used. The maximum kinetic energy of photoelectrons
ⓐ. decreases to zero
ⓑ. increases
ⓒ. remains fixed for all frequencies above \( \nu_0 \)
ⓓ. becomes equal to the work function
Correct Answer: increases
Explanation: For a given metal, the work function \( \phi_0 \) is fixed under the same surface condition. A higher incident frequency means a higher photon energy \(h\nu\). After the work function is used to release an electron, the remaining energy appears as maximum kinetic energy. Therefore, increasing frequency above threshold increases \(K_{\max}\). This frequency dependence is one of the central results of the photoelectric effect.
110. With the same metal, the incident frequency is increased while intensity is adjusted so that the saturation current remains unchanged. The stopping potential
ⓐ. decreases to zero
ⓑ. remains unchanged because current is unchanged
ⓒ. becomes independent of electron charge
ⓓ. increases
Correct Answer: increases
Explanation: Stopping potential depends on maximum kinetic energy through \(K_{\max}=eV_0\). Increasing frequency increases photon energy \(h\nu\). For the same metal, the work function remains unchanged, so the excess energy after emission increases. Hence \(K_{\max}\) increases and the stopping potential increases. Keeping the saturation current unchanged only controls the number of emitted electrons per second, not their maximum energy.
111. The frequency of incident radiation is changed from \(6.0\times10^{14}\,Hz\) to \(8.0\times10^{14}\,Hz\) for the same metal. If \(h=6.6\times10^{-34}\,J\,s\), the increase in maximum kinetic energy is
ⓐ. \(1.32\times10^{-19}\,J\)
ⓑ. \(6.6\times10^{-20}\,J\)
ⓒ. \(3.96\times10^{-19}\,J\)
ⓓ. \(5.28\times10^{-19}\,J\)
Correct Answer: \(1.32\times10^{-19}\,J\)
Explanation: \( \textbf{Initial frequency:} \) \( \nu_1=6.0\times10^{14}\,Hz \).
\( \textbf{Final frequency:} \) \( \nu_2=8.0\times10^{14}\,Hz \).
\( \textbf{Planck constant:} \) \(h=6.6\times10^{-34}\,J\,s\).
\( \textbf{For the same metal:} \) Work function is unchanged.
\( \textbf{Change in maximum kinetic energy:} \)
\[
\Delta K_{\max}=h(\nu_2-\nu_1)
\]
\( \textbf{Frequency difference:} \)
\[
\nu_2-\nu_1=2.0\times10^{14}\,Hz
\]
\( \textbf{Substitution:} \)
\[
\Delta K_{\max}=(6.6\times10^{-34})(2.0\times10^{14})\,J
\]
\( \textbf{Multiplication:} \)
\[
6.6\times2.0=13.2
\]
\( \textbf{Power of ten:} \)
\[
10^{-34}\times10^{14}=10^{-20}
\]
\( \textbf{Final simplification:} \)
\[
\Delta K_{\max}=13.2\times10^{-20}\,J=1.32\times10^{-19}\,J
\]
\( \textbf{Final answer:} \) The increase in maximum kinetic energy is \(1.32\times10^{-19}\,J\).
112. A current-potential curve shifts so that its zero-current intercept moves from \(-1.0\,V\) to \(-2.0\,V\), while the saturation current remains nearly the same. The most suitable change in incident radiation is
ⓐ. intensity increased while frequency remained the same
ⓑ. higher frequency with adjusted intensity
ⓒ. frequency decreased below threshold
ⓓ. work function of the same clean surface became zero
Correct Answer: higher frequency with adjusted intensity
Explanation: The negative intercept of the current-potential graph gives the stopping potential magnitude. A shift from \(-1.0\,V\) to \(-2.0\,V\) means the stopping potential has increased. Since \(K_{\max}=eV_0\), the fastest photoelectrons have greater maximum kinetic energy. That is naturally caused by increasing the frequency of incident radiation. The nearly same saturation current suggests that the emitted-electron rate was kept almost unchanged, likely by adjusting intensity.
113. Threshold-frequency comparison uses radiations of frequencies \(0.8\nu_0\), \( \nu_0 \), and \(1.5\nu_0\) incident separately on a metal whose threshold frequency is \( \nu_0 \). The case with the largest maximum kinetic energy is
ⓐ. \(0.8\nu_0\)
ⓑ. \( \nu_0 \)
ⓒ. \(1.5\nu_0\)
ⓓ. all three give the same \(K_{\max}\)
Correct Answer: \(1.5\nu_0\)
Explanation: Radiation of frequency \(0.8\nu_0\) is below threshold, so it does not produce photoelectric emission. At \( \nu_0 \), emission is just possible and \(K_{\max}=0\). At \(1.5\nu_0\), the photon energy is greater than the work function. The excess energy appears as maximum kinetic energy of the emitted photoelectrons. Among the given frequencies, \(1.5\nu_0\) gives the largest \(K_{\max}\).
114. A \(K_{\max}\)-versus-frequency graph for a given metal is described below.
The graph is a straight line. It cuts the frequency axis at \( \nu=\nu_0 \), and \(K_{\max}\) increases for \( \nu\gt\nu_0 \).
The frequency-axis intercept represents
ⓐ. the saturation current of the tube
ⓑ. the threshold frequency of the metal
ⓒ. the stopping potential for all metals
ⓓ. the photon momentum at zero wavelength
Correct Answer: the threshold frequency of the metal
Explanation: The graph of \(K_{\max}\) versus \( \nu \) follows the relation \(K_{\max}=h\nu-\phi_0\). At the threshold frequency, the emitted electron has zero maximum kinetic energy. Therefore, the graph cuts the frequency axis at \( \nu=\nu_0 \). Below this frequency, photoemission does not occur. The intercept is a property of the metal surface, not a measure of saturation current.
115. A comparison of frequency effects is shown below.
| Change made for the same metal | Effect on \(K_{\max}\) | Effect on \(V_0\) |
|---|
| P. Frequency increased above \( \nu_0 \) | Increases | Increases |
| Q. Frequency reduced to exactly \( \nu_0 \) | Becomes \(0\) | Becomes \(0\) |
| R. Frequency reduced below \( \nu_0 \) | No photoemission | No stopping potential measured |
The assessment of the table is
ⓐ. only P is valid
ⓑ. only P and Q are valid
ⓒ. only Q and R are valid
ⓓ. P, Q and R are valid
Correct Answer: P, Q and R are valid
Explanation: For frequencies above threshold, increasing frequency increases photon energy and hence increases \(K_{\max}\). Since \(K_{\max}=eV_0\), the stopping potential also increases. At the threshold frequency, the photon energy is just equal to the work function, so \(K_{\max}=0\) and \(V_0=0\). Below threshold, no photoelectrons are emitted, so stopping potential is not meaningfully measured. The table correctly separates above-threshold, threshold, and below-threshold cases.
116. A claim says, “Increasing the frequency of light always increases photoelectric current.” A better statement is that increasing frequency above threshold certainly tends to increase
ⓐ. work function and electron charge
ⓑ. saturation current at every fixed intensity without exception
ⓒ. the rest mass of photons
ⓓ. maximum kinetic energy and stopping potential
Correct Answer: maximum kinetic energy and stopping potential
Explanation: Increasing frequency increases photon energy according to \(E=h\nu\). For the same metal, the work function remains the same, so the excess energy available to photoelectrons increases. This increases \(K_{\max}\) and therefore increases stopping potential through \(K_{\max}=eV_0\). Photoelectric current depends mainly on how many photoelectrons are emitted and collected per second. At fixed intensity, changing frequency also changes photon number rate, so current should not be treated as automatically increasing just because frequency increases.
117. Two photoelectric \(I\)-\(V\) curves for the same metal reach the same saturation current, but curve \(P\) cuts the negative potential axis farther from the origin than curve \(Q\). This most likely means that curve \(P\) was obtained with
ⓐ. higher frequency with adjusted intensity
ⓑ. greater intensity at the same frequency
ⓒ. lower frequency and greater work function
ⓓ. below-threshold radiation of greater intensity
Correct Answer: higher frequency with adjusted intensity
Explanation: The negative potential-axis intercept gives the stopping potential magnitude. If curve \(P\) cuts the negative axis farther from the origin, its stopping potential is larger. A larger stopping potential means a larger maximum kinetic energy because \(K_{\max}=eV_0\). For the same metal, a larger \(K_{\max}\) is caused by a larger incident frequency. The same saturation current means the collected electron rate is nearly the same, so the intensity must have been adjusted to keep the emission rate unchanged.
118. The graph description for a photoelectric tube is given below.
Curve \(X\) and curve \(Y\) are plotted between photoelectric current \(I\) and collector potential \(V\). Both curves have the same stopping potential. Curve \(Y\) reaches twice the saturation current of curve \(X\).
The best conclusion is that curve \(Y\) corresponds to
ⓐ. twice the frequency at the same intensity
ⓑ. twice the intensity at the same frequency
ⓒ. half the work function of the same metal
ⓓ. below-threshold radiation with longer exposure time
Correct Answer: twice the intensity at the same frequency
Explanation: The same stopping potential means the same \(K_{\max}\) for the fastest photoelectrons. For the same metal, this points to the same incident frequency because photon energy is unchanged. The saturation current depends on the number of photoelectrons collected per second. If curve \(Y\) has twice the saturation current, the emission rate is doubled. At fixed above-threshold frequency, that is explained by doubling the intensity.
119. The retarding region of a photoelectric \(I\)-\(V\) curve is the part where
ⓐ. the collector is negative and current falls under retardation
ⓑ. the collector is strongly positive and all emitted electrons are collected
ⓒ. the incident frequency is below threshold and saturation current is maximum
ⓓ. the metal work function becomes independent of material
Correct Answer: the collector is negative and current falls under retardation
Explanation: A retarding potential is applied when the collector is negative with respect to the emitter. Since photoelectrons are negatively charged, the negative collector repels them and reduces their chance of reaching the collector. As the retarding potential increases in magnitude, the current decreases. At the stopping potential, even the fastest photoelectrons are stopped and the current becomes zero. This region is different from the positive-potential saturation region, where collection becomes complete.
120. A photoelectric-current table is shown below.
| Change made | Expected graph change |
|---|
| P. Increase intensity at fixed \( \nu\gt\nu_0 \) | Saturation current increases |
| Q. Increase frequency at adjusted intensity | Stopping potential increases |
| R. Make collector more negative | Current decreases in the retarding region |
| S. Increase positive collector potential after saturation | Current keeps increasing linearly without limit |
The mismatched row is
ⓐ. P
ⓑ. Q
ⓒ. S
ⓓ. R
Correct Answer: S
Explanation: Row P is correct because higher intensity at fixed suitable frequency increases the emitted-electron rate and hence saturation current. Row Q is correct because higher frequency increases \(K_{\max}\), so the stopping potential increases. Row R is correct because a more negative collector retards more photoelectrons and reduces current. Row S is mismatched because after saturation nearly all emitted photoelectrons are already collected. Beyond saturation, increasing positive collector potential does not make the current increase linearly without limit.
