An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is
- 2
- 3
- 4
- 5
Answer
(3) 4
Solution
Maximum kinetic energy of photoelectrons = eV0 = 10 eV.
Energy of the incident photon E = 0 + eV0 = 2.75 + 10 = 12.75 eV
For a jump from n to 1 in hydrogen: E = 13.6 (1 - 1n2 ) eV
1 - 1n2 = 12.7513.6 = 0.9375
1n2 = 0.0625 = 116
n = 4
Answer: (3)