Question

The circumference of the second Bohr orbit of electron in hydrogen atom is $600nm$. The potential difference that must be applied between the plates so that the electrons have the de Broglie wavelength corresponding in this circumference is

• A. ${10}^{-5}V$
• B. $\frac{5}{3}\times {10}^{-5}V$
• C. $5\times {10}^{-5}V$
• D. $3\times {10}^{-5}V$

Answer 1

Answer: (B)

$\frac{5}{3}\times {10}^{-5}V$

For second Bohr orbit, $600nm=n\lambda =2\lambda \Rightarrow \lambda =3000A^o$

For electron $\lambda =\frac{12.27}{\sqrt{V}}{A}^o$

Voltage $V={\left(\frac{12.27}{3000}\right)}^2=\frac{150.553}{9}\times {10}^{-6}=1.67\times {10}^{-5}=\frac{5}{3}\times {10}^{-5}V$