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:
Answer & explanation
Correct answer: option 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
$v^2∝(\frac{1}{λ})$
$∴\frac{v_1}{v_2}=\sqrt{\frac{(\frac{1}{λ_1})}{(\frac{1}{λ_2})}}=\sqrt{\frac{λ_2}{λ_1}}=\sqrt{\frac{3λ/4}{λ}}$
$∴v_2=v(\frac{4}{3})^{1/2}$