TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Given,
Energy given to hydrogen atom in ground state =∆E
Let initial angular momentum be L.
Then, final angular momentum be L′=L+‌

h

2π

Since, ‌‌L=‌

nh

2π

. . . (i)
Angular momentum of ground state (n=1)
L=‌

1⋅h

2π

=‌

h

2π

Hence, from Eq. (i), L′=‌

h

2π

+‌

h

2π

=2⋅‌

h

2π

Final orbit level, nf=2
Now, electrons will transit from orbit n=1 to n=2
Since, ‌‌∆E=13.6Z2(‌

1

nf2

−‌

1

ni2

) . . .(ii)
where, Z= atomic number,
and E= transition energy of electron, while moving ni to nf and Z (hydrogen atom) =1
Substituting in Eq. (ii), we get
∆E‌‌=13.6×12(‌

1

12

−‌

1

22

)
⇒‌‌∆E‌‌=13.6(‌

1

1

−‌

1

4

)
‌‌=13.6×‌

3

4

‌‌=3.4×3=10.2eV