# NEET Questions Solved

NEET - 2011

An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function 2.75 eV. If the stopping potential of the photo-electron is 10V,the value of n is

(a)3                                    (b)4

(c)5                                    (d)2

$E=K{E}_{max}+W$

$=e{V}_{0}+W$

$=10+2.75$

$E=12.75eV$

Difference of 4 and 1 energy level is 12.75 eV. So, higher energy level is 4 to ground and excited state is n=3.

Difficulty Level:

• 29%
• 40%
• 18%
• 15%
Crack NEET with Online Course - Free Trial (Offer Valid Till September 17, 2019)