The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency v of the incident radiation is doubled. The work function of this metal is:

Question

The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency v of the incident radiation is doubled. The work function of this metal is: (A) ( hv /4) (B) ( hv /3) (C) ( hv /2) (D) (2 hv /3)

Tardigrade Admin · Accepted Answer

As we know, ( K.E 1/ K.E 2)=( h ( v )-φ o / h (2 v )-φ o ) =((1/2) mv 2/(1)2 m (2 v )2) (1/4)=(h v-φ0/2 h v-φ) ⇒ 4 h v-4 φ0=2 h v-φ0 2 hv =3 φ o (2 hv /3)=φ o

Q. The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency $v$ of the incident radiation is doubled. The work function of this metal is:

$\frac{h v}{4}$

$\frac{h v}{3}$

$\frac{h v}{2}$

$\frac{2 h v}{3}$

As we know, $\frac{K . E _{1}}{K . E _{2}} = \frac{h ( v ) - ϕ _{o}}{h ( 2 v ) - ϕ _{o}}$
$= \frac{\frac{1}{2} m v ^{2}}{\frac{1}{2} m ( 2 v ) ^{2}}$
$\frac{1}{4} = \frac{h v - ϕ _{0}}{2 h v - ϕ}$
$\Rightarrow 4 h v - 4 ϕ_{0} = 2 h v - ϕ_{0}$
$2 h v = 3 ϕ_{o}$
$\frac{2 h v}{3} = ϕ_{o}$

Q. The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency v of the incident radiation is doubled. The work function of this metal is:

4hv​

3hv​

2hv​

32hv​