Question

The wavelength associated with an electron accelerated through a potential difference of 100 V is nearly

• A. 100 Å
• B. 123 Å
• C. 1.23 Å
• D. 0.123 Å

Answer 1

The correct option is C 1.23 Å

$\lambda =\frac{h}{\sqrt{2mQV}}=\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times 100}}=1.23\AA$