In the first excited state of hydrogen atom, the energy of its electron is −3.4 eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately : (Take 1 eV = 1.6 10-19 J, e = 1.6 10-19 C and 14 0 = 9 109 N m2/C2)
- 2.1 10-8 m
- 2.1 10-10 m
- 2.1 10-11 m
- 2.1 10-9 m
Answer
(2) 2.1 10-10 m
Solution
Total energy of the electron in a Bohr orbit: E = - ke22r, where k = 14 0
⇒ r = ke22|E| = 9 109 (1.6 10-19)22 3.4 1.6 10-19
⇒ r = 9 109 1.6 10-196.8
∴ r 2.1 10-10 m
Answer: (2)