Atomic Structure Question 9
Question: The energy of electron in first Bohr’s orbit of H-atom is -13.6 eV. What will be its potential energy in n = $ 4^{th} $ orbit
Options:
A) -13.6eV
B) -3.4eV
C) -0.85eV
D) -1.70eV
Show Answer
Answer:
Correct Answer: D
Solution:
- Energy of nth orbit of H-atom $ =-\frac{2{{\pi }^{2}}me^{4}k^{2}}{n^{2}h^{2}} $ Energy of Bohr’s orbit of H-atom $ =-\frac{2{{\pi }^{2}}me^{4}k^{2}}{h^{2}} $
$ =-13.6eV $ (given) Energy of fourth Bohr’s of H-atom $ =-\frac{2{{\pi }^{2}}me^{4}}{h^{2}n^{2}} $
$ =13.6\times \frac{1}{16}eV=-0.85eV $ PE of electron in nth orbit = $ 2\times E_{n} $ So P.E. of electron in $ 4^{th} $ orbit $ =2\times (-0.85)=-1.70eV $
BETA
