Question
An electron in the hydrogen atom jumps from excited state n to the ground state. The wavlength so emitted illuminates a photo -sensitive material having work function 2.75 eV. If the stopping potential of the photo-electron is 10 V, the value of n is
-
3
-
4
-
5
-
2
Solution
A.
3
E = KEmax + W
eVo + W
= 10 + 2.75
E = 12.75 eV
The difference of 4 and 1 energy level is 12.75 eV. So, the higher energy level is 4 to the ground and excited state is n=3.
