A photosensitive metallic surface has work function $\phi$. If photon of energy $3\phi$ fall on this surface, the electron comes out with a maximum velocity of $6×10^6ms^{-1}$. When the photon energy is increased to $9\phi$, there maximum velocity of photoelectron will be:
Answer & explanation
Correct answer: option 1
$ \text{Initially } K.E_{max} = \frac{1}{2} mv^2 = h\nu - \phi$
For Energy of $3\phi , K_1 = \frac{1}{2} mv_1^2 = 3\phi - \phi = 2\phi$
For Energy of $9\phi , K_2 = \frac{1}{2} mv_2^2 = 9\phi - \phi = 8\phi$
Divide these two we have $\Rightarrow \frac{v_2^2}{v_1^2} = 4$
$ \Rightarrow v_2 = 2v_1 = 12\times 10^6 m/s$