An $\alpha$-particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de Broglie wavelengths are $\lambda_{\alpha}$ and $\lambda_{p}$ respectively. The ratio $\frac{\lambda_{p}}{\lambda_{\alpha}}$, to the nearest integer, is

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Modern Physics

α

-particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de Broglie wavelengths are

λ_{α}

and

λ_{p}

respectively. The ratio

\frac{λ _{p}}{λ _{α}}

, to the nearest integer, is

Answer: 3

Solution

The de Broglie wavelength is

λ = \frac{h}{p} = \frac{h}{2 m K}

where

K

is kinetic energy.

After acceleration through potential V:

K = q V

.

For the proton:

λ_{p} = \frac{h}{2 m _{p} \cdot e V}

For the α-particle (charge

2 e

, mass

\approx 4 m_{p}

λ_{α} = \frac{h}{2 ( 4 m _{p} ) \cdot 2 e V} = \frac{h}{16 m _{p} e V}

Ratio:

\frac{λ _{p}}{λ _{α}} = \frac{16 m _{p} e V}{2 m _{p} e V} = \frac{16}{2} = 8 = 22 \approx 2.83

To the nearest integer: 3

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer

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