Question

Lights of two different frequencies whose photons have energies 1 and 2.5 eV, respectively, successively illuminate a metal whose work function is 0.5 eV. The ratio of the maximum speeds of the emitted electrons will be :

• A. 1:5
• B. 1:4
• C. 1:2
• D. 1:1

Answer 1

The correct option is C 1:2

From Einstein's equation for photoelectric effect , the maximum kinetic energy , $K=h\nu -W$

where $\nu =$ frequency of incident light , $h=$ Planck's constant and $W=$ work function of metal.

We know that energy of photons is also $E=h\nu$

so, $\frac{1}{2}m{v}^2=E-W$

Thus, $\frac{1}{2}m{v}_1^2=1-0.5=0.5....\left(1\right)$ and $\frac{1}{2}m{v}_2^2=2.5-0.5=2....\left(2\right)$

So, $\left(1\right)/\left(2\right)\Rightarrow \frac{v_1^2}{v_2^2}=\frac{0.5}{2}=1/4$

or $\frac{v_1}{v_2}=\frac{1}{2}$