Question

The velocity of the most energetic electrons emitted from a metal surface is doubled, when the frequency $\nu$ of 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. zero

Answer 1

Answer: (A)

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

$K.E{.}_{max}=h\nu -\varphi$

$\frac{1}{2}m{v}^2=h\nu -\varphi ------1$

$\frac{1}{2}(2v{)}^2=h2\nu -\varphi ------2$

$4\left(\frac{1}{2}m{v}^2\right)=h2\nu -\varphi ------3$

So, from $1$ put value of $\frac{1}{2}m{v}^2$ in equation $3$

$4(h\nu -\varphi )=h2\nu -\varphi$

$4h\nu -4\varphi =2h\nu -\varphi$

$2h\nu =3\varphi$

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