De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)
- 1.67 nm
- 2.67 nm
- 0.067 nm
- 0.67 nm
Answer
(4) 0.67 nm
Solution
Radius of the nth orbit: rn = n2a0 = 4 0.052 = 0.208 nm
Bohr's condition 2 rn = n gives = 2 rnn
= 2 3.14 0.2082 = 0.653 nm
∴ is close to 0.67 nm.
Answer: (4)