If p is the momentum of the fastest electron ejected from a metal surface after irradiation of light having wavelength…

🧠 The High-Energy Approximation When Kinetic Energy ($KE$) is much greater than the work function ($\Phi$), we can assume the incident photon's energy ($hc/\lambda$) is almost entirely converted into the electron's motion. This simplifies the math to a direct power-law ratio.

🗺️ The Momentum Scaling

1. The Proportionality: $KE = \frac{p^2}{2m} \approx \frac{hc}{\lambda}$ From this, we see that $\lambda \propto \frac{1}{p^2}$.
2. Setup the Ratio: $\frac{\lambda'}{\lambda} = \left( \frac{p}{p'} \right)^2$
3. Plugging in $1.5p$: $\frac{\lambda'}{\lambda} = \left( \frac{p}{1.5p} \right)^2 = \left( \frac{1}{1.5} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$ $\lambda' = \frac{4}{9}\lambda$

⚡ The Inverse Square Rule If momentum increases by a factor of $k$, the required wavelength must decrease by a factor of $k^2$. Since $1.5^2 = 2.25$ (which is $9/4$), the new wavelength is $1/2.25$ or $4/9$ of the original.

⚠️ Common Traps Don't use a linear inverse ratio ($1/k$). Momentum is linked to energy through its square ($p^2$), making the wavelength shift much steeper!

$\boxed{\text{Answer: (a)}}$