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Question: The work function for a metal is \(40\,{\rm{ev}}\). To emit photoelectrons of zero velocity from the...

The work function for a metal is 40ev40\,{\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.

Explanation

Solution

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=hv0+12mv2hv = h{v^0} + \dfrac{1}{2}m{v^2}, where hv is energy of incident photons, hv0h{v^0} or W is the work function and 12mv2\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=hv0+12mv2hv = h{v^0} + \dfrac{1}{2}m{v^2} ….. (1)

Now, come to the question. The work function is given as 40ev40\,{\rm{ev}}. And the velocity of the emitted photon is 0, that means, the kinetic energy of the emitted photon is zero.
KE=12mV2KE = \dfrac{1}{2}m{V^2}
As, velocity is zero, the above expression becomes,
KE=0KE = 0
Now, equation (1) becomes,

hv=hv0(W)hv = h{v^0}\left( W \right) …… (2)
The relation between v (frequency) and λ\lambda (wavelength) is,

v=cλv = \dfrac{c}{\lambda }

Substitute the value of v in equation (2).

hcλ=Wh\dfrac{c}{\lambda } = W …… (3)

Now, we have to convert 40eV to V.

40eV=40×1.6×1019=6.4×1018V \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, 3.0×108  m/s3.0 \times {10^8}\;m/{\mathop{\rm s}\nolimits} and h is the Planck’s constant whose value is 6.626×1034  m2kg/sec6.626 \times {10^{ - 34}}\;{{\rm{m}}^{\rm{2}}}{\rm{kg/sec}}. Now, we have to put the above values in equation (3).

hcλ=Wh\dfrac{c}{\lambda } = W

6.626×1034×3.0×108λ=6.4×1018 \Rightarrow 6.626 \times {10^{ - 34}} \times \dfrac{{3.0 \times {{10}^8}}}{\lambda } = 6.4 \times {10^{ - 18}}

19.878×1026×1λ=6.4×1018 \Rightarrow 19.878 \times {10^{ - 26}} \times \dfrac{1}{\lambda } = 6.4 \times {10^{ - 18}}

λ=3.106×108m \Rightarrow \lambda = 3.106 \times {10^{ - 8}}\,{\rm{m}}

We know that, 1nm=109{10^{ - 9}}m. So, value of λ\lambda is,

λ=31.06×109m=31.06nm \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).