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A hydrogen like atom with atomic number ...

A hydrogen like atom with atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition ot quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is – 13. 6 eV.

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(2n) -E_(1)=204 eV rArr (E_(1))/(4n^(2))-E_(1)=204 eV " "...(i)`
and `E_(2n) -E_(n) = 40.8 eV rArr (E_(1))/(4n^(2))-(E_(1))/(n^(2))= 40.8 eV " "...(ii)`
From (i)/(ii)
`(1-1/(4n^(2)))/(3/(4n^(2)))=5 rArr n = 2 `.
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