Photoelectric Effect
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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357657 When a metal surface is illuminated by a monochromatic light of wave length \(\lambda\), then the stopping potential difference required is \(3\;V\). When the same surface is illuminated by the light of wavelength \(2 \lambda\), then the stopping potential required is \(V\). Then the threshold wavelength of the given metal surface will be

1 \(6 \lambda\)
2 \(4 \lambda / 3\)
3 \(4 \lambda\)
4 \(8 \lambda\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357658 The threshold frequency of a metal with workfunction \(6.63\,\,eV\) is:

1 \(1.6 \times {10^{12}}\,\;Hz\)
2 \(16 \times {10^{12}}\,\;Hz\)
3 \(16 \times {10^{15}}\,\;Hz\)
4 \(1.6 \times {10^{15}}\,\;Hz\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357659 Assertion :
Stopping potential is a measure of \(KE\) of photoelectron.
Reason :
\(W=e V_{s}=\dfrac{1}{2} m v^{2}=K E\)

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357660 The threshold frequency of a metal is \(f_{0}\). When the light of frequency \(2 f_{0}\) is incident on the metal plate, the maximum velocity of photoelectrons is \(v_{1}\). When the frequency of incident radiation is increased to \(5 f_{0}\), the maximum velocity of photoelectrons emitted is \(v_{2}\). The ratio of \(v_{1}\) to \(v_{2}\) is

1 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{4}\)
2 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{16}\)
3 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{8}\)
4 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{2}\)
JEE - 2023
NEET DIGITAL LIBRARY
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357657 When a metal surface is illuminated by a monochromatic light of wave length \(\lambda\), then the stopping potential difference required is \(3\;V\). When the same surface is illuminated by the light of wavelength \(2 \lambda\), then the stopping potential required is \(V\). Then the threshold wavelength of the given metal surface will be

1 \(6 \lambda\)
2 \(4 \lambda / 3\)
3 \(4 \lambda\)
4 \(8 \lambda\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357658 The threshold frequency of a metal with workfunction \(6.63\,\,eV\) is:

1 \(1.6 \times {10^{12}}\,\;Hz\)
2 \(16 \times {10^{12}}\,\;Hz\)
3 \(16 \times {10^{15}}\,\;Hz\)
4 \(1.6 \times {10^{15}}\,\;Hz\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357659 Assertion :
Stopping potential is a measure of \(KE\) of photoelectron.
Reason :
\(W=e V_{s}=\dfrac{1}{2} m v^{2}=K E\)

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357660 The threshold frequency of a metal is \(f_{0}\). When the light of frequency \(2 f_{0}\) is incident on the metal plate, the maximum velocity of photoelectrons is \(v_{1}\). When the frequency of incident radiation is increased to \(5 f_{0}\), the maximum velocity of photoelectrons emitted is \(v_{2}\). The ratio of \(v_{1}\) to \(v_{2}\) is

1 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{4}\)
2 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{16}\)
3 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{8}\)
4 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{2}\)
JEE - 2023
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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357657 When a metal surface is illuminated by a monochromatic light of wave length \(\lambda\), then the stopping potential difference required is \(3\;V\). When the same surface is illuminated by the light of wavelength \(2 \lambda\), then the stopping potential required is \(V\). Then the threshold wavelength of the given metal surface will be

1 \(6 \lambda\)
2 \(4 \lambda / 3\)
3 \(4 \lambda\)
4 \(8 \lambda\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357658 The threshold frequency of a metal with workfunction \(6.63\,\,eV\) is:

1 \(1.6 \times {10^{12}}\,\;Hz\)
2 \(16 \times {10^{12}}\,\;Hz\)
3 \(16 \times {10^{15}}\,\;Hz\)
4 \(1.6 \times {10^{15}}\,\;Hz\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357659 Assertion :
Stopping potential is a measure of \(KE\) of photoelectron.
Reason :
\(W=e V_{s}=\dfrac{1}{2} m v^{2}=K E\)

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357660 The threshold frequency of a metal is \(f_{0}\). When the light of frequency \(2 f_{0}\) is incident on the metal plate, the maximum velocity of photoelectrons is \(v_{1}\). When the frequency of incident radiation is increased to \(5 f_{0}\), the maximum velocity of photoelectrons emitted is \(v_{2}\). The ratio of \(v_{1}\) to \(v_{2}\) is

1 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{4}\)
2 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{16}\)
3 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{8}\)
4 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{2}\)
JEE - 2023
For More Updates join with Us
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357657 When a metal surface is illuminated by a monochromatic light of wave length \(\lambda\), then the stopping potential difference required is \(3\;V\). When the same surface is illuminated by the light of wavelength \(2 \lambda\), then the stopping potential required is \(V\). Then the threshold wavelength of the given metal surface will be

1 \(6 \lambda\)
2 \(4 \lambda / 3\)
3 \(4 \lambda\)
4 \(8 \lambda\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357658 The threshold frequency of a metal with workfunction \(6.63\,\,eV\) is:

1 \(1.6 \times {10^{12}}\,\;Hz\)
2 \(16 \times {10^{12}}\,\;Hz\)
3 \(16 \times {10^{15}}\,\;Hz\)
4 \(1.6 \times {10^{15}}\,\;Hz\)
JEE - 2024
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357659 Assertion :
Stopping potential is a measure of \(KE\) of photoelectron.
Reason :
\(W=e V_{s}=\dfrac{1}{2} m v^{2}=K E\)

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357660 The threshold frequency of a metal is \(f_{0}\). When the light of frequency \(2 f_{0}\) is incident on the metal plate, the maximum velocity of photoelectrons is \(v_{1}\). When the frequency of incident radiation is increased to \(5 f_{0}\), the maximum velocity of photoelectrons emitted is \(v_{2}\). The ratio of \(v_{1}\) to \(v_{2}\) is

1 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{4}\)
2 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{16}\)
3 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{8}\)
4 \(\dfrac{v_{1}}{v_{2}}=\dfrac{1}{2}\)
JEE - 2023
NEET DIGITAL LIBRARY