$$K_{\max} = \frac{hc}{\lambda} - \phi = \frac{12400}{4000} - 2 = 3.1 - 2 = \boxed{1.1 \text{ eV}}$$
Back to NEET UG 2025 Modern Physics
NEET UG 2025 — Physics Modern Physics
easy
mcq
2025
Official previous-year question
Verified 30 May 2026.
Question
The maximum kinetic energy of photoelectrons from a surface with work function $\phi = 2 \text{ eV}$ when light of $\lambda = 4000 \text{ Å}$ falls on it is:
$(hc = 12400 \text{ eV·Å})$
Options
- A
$1.1 \text{ eV}$
- B
$2 \text{ eV}$
- C
$3.1 \text{ eV}$
- D
$0.5 \text{ eV}$
Solution
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.