Question

The de broglie wavelength $\left( \lambda \right)$associated with a photoelectron varies with the frequency $\left( \gamma \right)$of the incident radiation as [${\upsilon _o}$is threshold frequency]:
(A).$\lambda \alpha \dfrac{1}{{{{\left( {\upsilon - {\upsilon _o}} \right)}^{\dfrac{3}{2}}}}}$
(B).$\lambda \alpha \dfrac{1}{{{{\left( {\upsilon - {\upsilon _o}} \right)}^{\dfrac{1}{2}}}}}$
(C).$a\lambda \alpha \dfrac{1}{{{{\left( {\upsilon - {\upsilon _0}} \right)}^{\dfrac{1}{4}}}}}$
(D).$\lambda \alpha \dfrac{1}{{\left( {\upsilon - {\upsilon _o}} \right)}}$

Answer 1

Hint:The wavelength associated with de-Broglie radiations is known as se-Broglie wavelength threshold frequency refers to the minimum frequency required to remove electrons from the metal surface.

Complete step by step answer:
As we know that radiation has dual nature i.e. it passes properties of both wave and particle nature. Therefore de-Broglie concluded that the moving material particle must also pass dual nature, since nature loves symmetry since one put forward the de-Broglie hypothesis. According to de-Broglie a moving material particle sometime as a particle which controls the particle in every respect, the wave associated with moving particle is called matter wave as de-Broglie wave whose wavelength is called de-Broglie wavelength is given by $\lambda = \dfrac{h}{{mv}}....(1)$.
we know that kinetic energy (KE) =$\dfrac{1}{2}m{v^2}$
$ar$$v = \sqrt {\dfrac{{2(KE)}}{m}} ......(2)$
On putting value of equation (2) in (1), we get
$\lambda = \dfrac{\lambda }{{\sqrt {2m(KE)} }}.....(3)$
Now according to the Einstein photoelectric equation, we know that kinetic energy (KE) $h\left( {v - {v_o}} \right)...(4)$.
Here $h$ is Planck’s constant, $v$ is frequency of incident radiation and ${v_o}$ is the threshold frequency
Put value of KE from quotation (4) in eq. (3)
$\lambda = \dfrac{h}{{\sqrt {2mh\left( {v - {v_o}} \right)} }}$ or
$\lambda = \sqrt {\dfrac{h}{{2m}}} \times \sqrt {\dfrac{l}{{v - {v_o}}}} $
$\lambda \alpha \dfrac{l}{{\sqrt {v - {v_o}} }}$ $\lambda \alpha \dfrac{l}{{{{\left( {v - {v_o}} \right)}^{\dfrac{1}{2}}}}}$
Hence we get the required relation.
Hence the correct option is B.

Note:
We have taken $\sqrt {\dfrac{h}{{2m}}}$ as a constant here because we know that $h$ is a Planck constant whose value is constant lawyers. Merely the mass of a particle is fixed at attaining condition so we have taken it as a constant.