An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength 
, energy E = 
):
a)
n = 5
b)
n = 7
c)
n = 4
d)
n = 6
, energy E = ):a)
n = 5
b)
n = 7
c)
n = 4
d)
n = 6
Question ID - 51449 | SaraNextGen Top Answer An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength , energy E = ):
a)
n = 5
b)
n = 7
c)
n = 4
d)
n = 6
An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength , energy E = ):
|
a) |
n = 5 |
b) |
n = 7 |
|
c) |
n = 4 |
d) |
n = 6 |
1 Answer
127 votes
Answer Key / Explanation : (a) -
⇒ 
Let it start from n to m and from m to ground.
Then 13.6 
4
= 
⇒ 
= 0.7498
⇒ 0.25 = 
m = 2, and now 13.6 
4
= 
n = 5.
127 votes
Answer Link
127