Question

De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to: $(|e|=1.6\times {10}^{-19}C,m_e=9.1\times {10}^{-31}kg,h=6.6\times {10}^{-34}Js).$

• A. $0.5\mathring{A}$
• B. $1.7\mathring{A}$
• C. $2.4\mathring{A}$
• D. $1.2\mathring{A}$

Answer 1

The correct option is B $1.7\mathring{A}$

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2mqV}}$

$\lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times 50}}=1.72\times {10}^{-10}=1.72A$

Option B