JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Atomic Structure

Because the kinetic energy of the ejected electron is taken to be much larger than the work function, the photon energy goes almost entirely into kinetic energy, $\frac{hc}{\lambda }\approx \text{KE}=\frac{p^2}{2m}$. So the photon energy, and hence $\frac{1}{\lambda }$, is proportional to $p^2$. Going from momentum $p$ to $1.5p$ multiplies the kinetic energy by $1.5^2=2.25=\frac{9}{4}$, so $\frac{1}{\lambda }$ must grow by that same factor. The wavelength therefore shrinks to $\frac{4}{9}$ of its original value, $\frac{4}{9}\lambda$.