Question

The velocity of electron of H-atom in its ground state is $2.2\times {10}^{-6}m/s$. The de-Broglie wavelength of this electron would be:

• A. $0.33nm$
• B. $23.3nm$
• C. $45.6nm$
• D. $100nm$

Answer 1

Answer: (A)

$0.33nm$

Given:

$v=2.2\times {10}^{-6}m/s,m_e=9.1\times {10}^{31},h=6.61\times {10}^{-34}$

$\lambda =\frac{h}{mv}=\frac{6.626\times {10}^{-34}}{9.1\times {10}^{-31}\times 2.2\times {10}^{-6}}=0.33\times {10}^{-9}=0.33nm$

Hence, the correct option is A.