An alpha-particle and a proton are accelerated from rest by a potential difference of 100 V- After this- their de Broglie wavelengths are lambda-alpha- and lambda-p- respectively- The ratio dfrac-lambda-p-lambda-alpha- to the nearest integer- is-

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We have- 12mv2-eVDe broglie wavelength- -x3BB-hmv-x3BB-h-x221A-2meV-x3BB-p-h-x221A-2-m-eV-1-x3BB-x3B1-h2-4m-2eV-2-Now- -x3BB-p-x3BB-x3B1-x221A-8-x2243-3

An $α$ -particle and a proton are accelerated from rest by a potential difference of $100 V$ . 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:

We have, $\frac{1}{2} m v^{2} = e V$

De broglie wavelength= $λ = \frac{h}{m v}$

$λ = \frac{h}{\sqrt{2 m e V}}$

$λ_{p} = \frac{h}{\sqrt{2 (m) e V}}$ ....(1)

$λ_{α} = \frac{h}{2 (4 m) 2 e V}$ ....(2)

Now, $\frac{λ_{p}}{λ_{α}} = \sqrt{8} ≃ 3$

An α-particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de Broglie wavelengths are λα and λp respectively. The ratio λpλα, to the nearest integer, is:

We have, 12mv2=eVDe broglie wavelength= λ=hmvλ=h√2meVλp=h√2(m)eV....(1)λα=h2(4m)2eV....(2)Now, λpλα=√8≃3

An $α$ -particle and a proton are accelerated from rest by a potential difference of $100 V$ . 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:

We have, $\frac{1}{2} m v^{2} = e V$

De broglie wavelength= $λ = \frac{h}{m v}$

$λ = \frac{h}{\sqrt{2 m e V}}$

$λ_{p} = \frac{h}{\sqrt{2 (m) e V}}$ ....(1)

$λ_{α} = \frac{h}{2 (4 m) 2 e V}$ ....(2)

Now, $\frac{λ_{p}}{λ_{α}} = \sqrt{8} ≃ 3$