The stopping potential for photoelect ons from a metal surface is V1 when monochromatic light of frequency v1 is incident on it. The stopping potential becomes V2 when monochromatic light of another frequency is incident on the same metal surface. If h be the Planck’s constant and e be the charge of an electron, then the frequency of light in the second case is

Question

The stopping potential for photoelect ons from a metal surface is V1 when monochromatic light of frequency v1 is incident on it. The stopping potential becomes V2 when monochromatic light of another frequency is incident on the same metal surface. If h be the Planck’s constant and e be the charge of an electron, then the frequency of light in the second case is (A) v1-(e/h)(V2+V1) (B) v1+(e/h)(V2+V1) (C) v1-(e/h)(V2-V1) (D) v1+(e/h)(V2-V1)

Tardigrade Admin · Accepted Answer

Here, h v1=φ0=e V1 .... (i) and h v2=φ0+e V2..... (ii) From Eqs. (i) and (ii), we have h( v 2- v 1)= e (V2-V1) ⇒ v2=(e/h)(V2-V1)+v1

$v_{1} - \frac{e}{h} (V_{2} + V_{1})$

$v_{1} + \frac{e}{h} (V_{2} + V_{1})$

$v_{1} - \frac{e}{h} (V_{2} - V_{1})$

$v_{1} + \frac{e}{h} (V_{2} - V_{1})$

Here, $h v_{1} = ϕ_{0} = e V_{1} .... (i)$
and $h v_{2} = ϕ_{0} + e V_{2}$ ..... (ii)
From Eqs. (i) and (ii), we have $h (v_{2} - v_{1}) = e (V_{2} - V_{1})$
$\Rightarrow v_{2} = \frac{e}{h} (V_{2} - V_{1}) + v_{1}$

v1​−he​(V2​+V1​)

v1​+he​(V2​+V1​)

v1​−he​(V2​−V1​)

v1​+he​(V2​−V1​)

Here, hv1​=ϕ0​=eV1​....(i) and hv2​=ϕ0​+eV2​..... (ii) From Eqs. (i) and (ii), we have h(v2​−v1​)=e(V2​−V1​) ⇒v2​=he​(V2​−V1​)+v1​

$v_{1} - \frac{e}{h} (V_{2} + V_{1})$

$v_{1} + \frac{e}{h} (V_{2} + V_{1})$

$v_{1} - \frac{e}{h} (V_{2} - V_{1})$

$v_{1} + \frac{e}{h} (V_{2} - V_{1})$

Here, $h v_{1} = ϕ_{0} = e V_{1} .... (i)$
and $h v_{2} = ϕ_{0} + e V_{2}$ ..... (ii)
From Eqs. (i) and (ii), we have $h (v_{2} - v_{1}) = e (V_{2} - V_{1})$
$\Rightarrow v_{2} = \frac{e}{h} (V_{2} - V_{1}) + v_{1}$