A photoemissive substance is illuminated with a radiation of wavelength λ i so that it releases electrons with de-Broglie wavelength λ e . The longest wavelength of radiation that can emit photoel

Question

A photoemissive substance is illuminated with a radiation of wavelength λ i ​ so that it releases electrons with de-Broglie wavelength λ e ​ . The longest wavelength of radiation that can emit photoelectron is λ 0 ​ . Expression for de-Broglie wavelength is (m = mass of electron, h = Planck's constant, C = Speed of light)

TestHub · Accepted Answer

Using Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is given by:

$K_{max} = \frac{h c}{λ _{i}} - \frac{h c}{λ _{0}} = h c (\frac{1}{λ _{i}} - \frac{1}{λ _{0}})$

The de-Broglie wavelength of the emitted electron is:

$λ_{e} = \frac{h}{p} = \frac{h}{2 m K _{max}}$

Substituting the value of $K_{max}$ :

$λ_{e} = \frac{h}{2 mh c ( \frac{1}{λ _{i}} - \frac{1}{λ _{0}} )}$

$λ_{e} = \frac{h ^{2}}{2 mh c ( \frac{1}{λ _{i}} - \frac{1}{λ _{0}} )}$

$λ_{e} = \frac{h}{2 m c ( \frac{1}{λ _{i}} - \frac{1}{λ _{0}} )}^{\frac{1}{2}}$

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