Question

The velocity by of the most energetic electrons emitted from a metallic surface is doubled when the frequency $\nu$ of the incident radiation is doubled. The work function of the metal is-

• A. $\frac{2}{3}h\nu$
• B. $\frac{h\nu }{2}$
• C. $\frac{h\nu }{3}$
• D. $\frac{h\nu }{4}$

Answer 1

The correct option is A $\frac{2}{3}h\nu$

According to Einstein's photo electric equation,

$E={\varphi }_0+K{E}_{max}$

Where, ${\varphi }_0=\text{Work function of the material}$

$h\nu ={\varphi }_0+\frac{1}{2}m{v}^2.......\left(1\right)$

When incident frequency is doubled, the velocity of emitted electrons also doubled, i.e.,

$h\left(2\nu \right)={\varphi }_0+\frac{1}{2}m(2v{)}^2$

$2h\nu ={\varphi }_0+4\left(\frac{1}{2}m{v}^2\right)......\left(2\right)$

From (1) we get,

$\frac{1}{2}m{v}^2=h\nu -{\varphi }_0......\left(3\right)$

From (2) $\&$ (3) we get,

$2h\nu ={\varphi }_0+4[h\nu -{\varphi }_0]$

$3{\varphi }_0=2h\nu$

${\varphi }_0=\frac{2}{3}h\nu$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.