A proton accelerated through a potential difference of 100V, has de – Broglie wavelength λ0 . The de – Broglie wavelength of an α - particle, accelerated through 800 V is

A
$\frac{λ_{0}}{\sqrt{2}}$
B
$\frac{λ_{0}}{2}$
C
$\frac{λ_{0}}{4}$
D
$\frac{λ_{0}}{8}$

Solution:

$\frac{p^{2}}{2 m} = qV$
$\Rightarrow p = \sqrt{2 mqV}$
$\Rightarrow λ = \frac{h}{\sqrt{2 mqV}}$
$\Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{m_{2} q_{2} V_{2}}{m_{1} q_{1} V_{1}}} = \sqrt{\frac{4 m \times 2 q \times 800}{m \times q \times 100}}$
$\Rightarrow \frac{λ_{0}}{λ_{2}} = \sqrt{64}$
$\Rightarrow λ_{2} = \frac{λ_{o}}{8}$

A proton accelerated through a potential difference of 100V, has de – Broglie wavelength $λ_{0}$ . The de – Broglie wavelength of an $α$ - particle, accelerated through 800 V is

Solution:

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