Question

The work function for a metal is $40\,{\rm{ev}}$. To emit photoelectrons of zero velocity from the surface of the metal the wavelength of incident light should be x nm. Find the value of x.

Answer 1

Hint:We know that, work function for a metal is the minimum energy needed to eject an electron from the surface of a metal. Here, we have to use the formula, $hv = h{v^0} + \dfrac{1}{2}m{v^2}$, where hv is energy of incident photons, $h{v^0}$ or W is the work function and $\dfrac{1}{2}m{v^2}$ is the kinetic energy of emitted electron.

Complete step by step answer:Let’s first discuss the photoelectric effect. This effect is the phenomenon of ejecting the electrons from the metallic surface under the influence of striking photons.
The photoelectric equation is,
$hv = h{v^0} + \dfrac{1}{2}m{v^2}$ ….. (1)

Now, come to the question. The work function is given as $40\,{\rm{ev}}$. And the velocity of the emitted photon is 0, that means, the kinetic energy of the emitted photon is zero.
$KE = \dfrac{1}{2}m{V^2}$
As, velocity is zero, the above expression becomes,
$KE = 0$
Now, equation (1) becomes,

$hv = h{v^0}\left( W \right)$ …… (2)
The relation between v (frequency) and $\lambda $ (wavelength) is,

$v = \dfrac{c}{\lambda }$

Substitute the value of v in equation (2).

$h\dfrac{c}{\lambda } = W$ …… (3)

Now, we have to convert 40eV to V.

$ \Rightarrow 40\,{\rm{eV}} = 40 \times 1.6 \times {10^{ - 19}} = 6.4 \times {10^{ - 18}}\,{\rm{V}}$

We know that, value of c is speed of light, that is, $3.0 \times {10^8}\;m/{\mathop{\rm s}\nolimits} $ and h is the Planck’s constant whose value is $6.626 \times {10^{ - 34}}\;{{\rm{m}}^{\rm{2}}}{\rm{kg/sec}}$. Now, we have to put the above values in equation (3).

$h\dfrac{c}{\lambda } = W$

$ \Rightarrow 6.626 \times {10^{ - 34}} \times \dfrac{{3.0 \times {{10}^8}}}{\lambda } = 6.4 \times {10^{ - 18}}$

$ \Rightarrow 19.878 \times {10^{ - 26}} \times \dfrac{1}{\lambda } = 6.4 \times {10^{ - 18}}$

$ \Rightarrow \lambda = 3.106 \times {10^{ - 8}}\,{\rm{m}}$

We know that, 1nm=${10^{ - 9}}$m. So, value of $\lambda $ is,

$ \Rightarrow \lambda = 31.06 \times {10^{ - 9}}\,{\rm{m}} = 31.06\,{\rm{nm}}$

Hence, the wavelength of light is 31.06 nm.


Note: The photoelectric effect can be explained with the help of particle nature which consists of a stream of photons. Certain binding energy holds electrons to the nucleus. In order to escape electrons, there must be a supply of energy to overcome the binding energy. This job is done by photons that must contain minimum energy (threshold energy).