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# The photoelectric threshold wavelength of silver is $3250 \times {10^{ - 19}}m$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times {10^{ - 10}}m$is:(Given $h = 4.14 \times {10^{ - 15}}eVs$ and $c = 3 \times {10^8}m{s^{ - 1}}$)(A) $\approx 0.3 \times {10^6}m{s^{ - 1}}$(B) $\approx 6 \times {10^5}m{s^{ - 1}}$(C) $\approx 6 \times {10^6}m{s^{ - 1}}$(D) $\approx 61 \times {10^3}m{s^{ - 1}}$

Last updated date: 17th Sep 2024
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Hint According to Einstein’s equation for the photoelectric effect, the kinetic energy of the electron is given by the difference in the incident energy and the wave function of the electron. And the kinetic energy is given by half of mass times the velocity squared. This can be used to determine the velocity of an electron.
Formula used:
$\Rightarrow$ $K = \dfrac{1}{2}m{v^2}$
$\Rightarrow$ $K = E - \phi$
$\Rightarrow$ $E = \dfrac{{hc}}{\lambda }$
Where $K$ is the kinetic energy of the electron.
$\Rightarrow$ $m$ is the mass of the electron.
$\Rightarrow$ $v$ is the velocity with which the electron is ejected.
$\Rightarrow$ $E$ is the energy of incident light.
$\Rightarrow$ $\phi$ is the work function of the electron.
$h$ is the Planck’s constant.
$\Rightarrow$ $c$ is the speed of light in vacuum.
$\lambda$ is the wavelength of the light.

Complete Step by step solution
When light with a frequency greater than the threshold frequency strikes photoelectric metal, some part of this energy called Work Function $\left( \phi \right)$ is used to provide enough energy to the electrons to escape the lattice, the rest of the energy is used as kinetic energy with which the electron travels away from the metal plate. This relation can be given by-
$\Rightarrow$ $E = K + \phi$
The kinetic energy is-
$\Rightarrow$ $K = E - \phi$
The threshold wavelength$\left( {{\lambda _0}} \right)$is the wavelength at which the electron gains enough energy to escape the lattice. The energy carried by the threshold frequency is equal to the work function of the electron.
Therefore,
$\Rightarrow$ $\phi = \dfrac{{hc}}{{{\lambda _0}}}$
The incident energy on the silver surface,
$\Rightarrow$ $E = \dfrac{{hc}}{\lambda }$
Therefore,
$\Rightarrow$ $K = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}$
We know that,
$K = \dfrac{1}{2}m{v^2}$
The equation becomes-
$\Rightarrow$ $\dfrac{1}{2}m{v^2} = hc\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right)$
Rearranging,
$\Rightarrow$ $v = \sqrt {\dfrac{{2hc}}{m}\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right)}$
It is given in the question that,
Threshold wavelength,${\lambda _0} = 3250 \times {10^{ - 19}}m$
Incident wavelength, $\lambda = 2536 \times {10^{ - 10}}m$
Velocity of light, $c = 3 \times {10^8}m{s^{ - 1}}$
Mass of electron, $m = 9.1 \times {10^{ - 31}}kg$
The Planck’s constant, $h = 4.14 \times {10^{ - 15}}eVs$
Converting this into SI units,
$\Rightarrow$ $1eVs = 1.602 \times {10^{ - 19}}Js$
$\Rightarrow$ $h = 4.14 \times 1.602 \times {10^{ - 15}} \times {10^{ - 19}}Js$
$\Rightarrow$ $h = 6.63 \times {10^{ - 34}}Js$
Putting these values in the equation,
$v = \sqrt {\dfrac{{2 \times 6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{9.1 \times {{10}^{ - 31}}}}\left( {\dfrac{1}{{2536 \times {{10}^{ - 10}}}} - \dfrac{1}{{3250 \times {{10}^{ - 19}}}}} \right)}$
$v = \sqrt {\dfrac{{2 \times 6.63 \times 3 \times {{10}^{ - 34}} \times {{10}^8}}}{{9.1 \times {{10}^{ - 31}} \times {{10}^{ - 29}}}}\left( {\dfrac{{3250 \times {{10}^{ - 19}} - 2536 \times {{10}^{ - 10}}}}{{2536 \times 3250}}} \right)}$
On solving this equation we get,
$v \approx 6 \times {10^5}m/s$

Therefore, option (B) is correct.

Note To obtain the velocity in the SI units, the value of Planck’s constant must be converted into SI units and the mass of the electron must also be written in kilograms. Failure in doing so may give incorrect answers. Also, the equation for velocity can be solved step by step instead of a large-single equation to avoid calculation mistakes.