The threshold frequency for a certain metal is $v_0$. When light of frequency $v = 2v_0$ is incident on it, the maximum velocity of photoelectrons is 4 × 10⁶ m/s. If the frequency of incident radiations is increased to $5v_0$, then the maximum velocity of photoelectrons in m/s will be -

$8×10^6$

$V_{max.}=\sqrt{\frac{2}{m}(E_{Ph}-W)}=\sqrt{\frac{2}{m}(hv-hv_0)}$

$(V_{max})_1=\sqrt{\frac{2h}{m}(2v_0-v_0)}$

$(V_{max})_2=\sqrt{\frac{2h}{m}(5v_0-v_0)}$

$\frac{V_{max._1}}{V_{max._2}}=2⇒V_{max._2}=2V_{max._1}$

$V_{max._2}=2×4×10^6=8×10^6m/s$