JEE Main 2015 · 10 Apr · Shift 1 · Physics Q27

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PhysicsSingle correct+4 / -1Question 27 of 90

de-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to ( $|e|=1.6\times10^{-19}$ C, $m_e=9.1\times10^{-31}$ kg, $h=6.6\times10^{-34}$ Js) :

0.5 Å

1.2 Å

1.7 Å

2.4 Å

Answer and worked solution

Official answer: C

$\lambda=\frac h{\sqrt{2meV}}=\frac{6.6\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times50}}\approx1.7\times10^{-10}\text{ m}=1.7\text{ Å}.$

Topics:de Broglie waves

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 27