In a photocell, with exciting wavelength λ, the faster electron has speed v. If the exciting wavelength is changed to 3λ/4, the speed of the fastest electron will be:

$v(\frac{4}{3})^{1/2}$

According to Einstein’s photoelectric equation,

$\frac{hc}{λ_1}=W_0+\frac{1}{2}mv_1^2$

and $\frac{hc}{λ_2}=W_0+\frac{1}{2}mv_2^2$

These expressions show that

$v^2∝(\frac{1}{λ})$

$∴\frac{v_1}{v_2}=\sqrt{\frac{(\frac{1}{λ_1})}{(\frac{1}{λ_2})}}=\sqrt{\frac{λ_2}{λ_1}}=\sqrt{\frac{3λ/4}{λ}}$

$∴v_2=v(\frac{4}{3})^{1/2}$