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. $\frac{ hv }{4}$
• B. $\frac{ hv }{3}$
• C. $\frac{ hv }{2}$
• D. $\frac{2 hv }{3}$

Answer 1

Answer: (D)

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