Explanation
The frequency of incident light is less than the threshold frequency. No photoelectrons will be emitted from the metal surface. The threshold frequency is $$7.24 \times 1014 Hz,$$ and no photoelectron will be ejected.
The energy of a neutron in eV whose de-Broglie wavelength is $$1\mathop {\text{A}}\limits^{\text{0}} $$
A non-monochromatic light is used in an experiment on the photoelectric effect. The stopping potential is related to the:
$$\begin{array}{l} Power\, of\, a\, bulb\left( p \right) =40W \\ energy\, emitted\, by\, a\, bulb=power\times Time\left( s \right) =40\times 20=800J \\ Energy\, of\, photons\, emitted\, by\, a\, bulb=\left( { \dfrac { { 80 } }{ { 100 } } } \right) \times 800=640J \\ Wavelength\left( \lambda \right) =620nm \\ Energy\, of\, a\, photon\, =\dfrac { { hc } }{ \lambda } =\dfrac { { 12400\times 1.6\times { { 10 }^{ -19 } }\times { { 10 }^{ -10 } } } }{ { 620\times { { 10 }^{ -9 } } } } \\ =\dfrac { { 640 } }{ { 3.2\times { { 10 }^{ -19 } } } } =2\times { 10^{ 21 } }\, photons \end{array}$$
Hence,
option $$(D)$$ is correct answer.
$$\begin{array}{l} \lambda =\dfrac { h }{ p } \, \, and\, \, \, \dfrac { { { p^{ 2 } } } }{ { 2m } } =\varepsilon =\dfrac { 3 }{ 2 } kT \\ \therefore p={ \left( { 3mkT } \right) ^{ \dfrac { 1 }{ 2 } } } \\ \lambda =\dfrac { h }{ p } =\dfrac { h }{ { { { \left( { 3kmT } \right) }^{ \dfrac { 1 }{ 2 } } } } } \\ \therefore T=\dfrac { { { h^{ 2 } } } }{ { 3km{ \lambda ^{ 2 } } } } \\ T=\dfrac { { { h^{ 2 } } } }{ { 3k{ \lambda ^{ 2 } } } } =\dfrac { { { { \left( { 6.626\times { { 10 }^{ -34 } } } \right) }^{ 2 } } } }{ { \left( 3 \right) \left( { 1.38\times { { 10 }^{ -23 } } } \right) \left( { 9.108\times { { 10 }^{ -31 } } } \right) { { \left( { 76.3\times { { 10 }^{ -9 } } } \right) }^{ 2 } } } } =2.00K \\ T=2.00K \end{array}$$
Option $$B$$ is correct .
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