A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of $50 V$ . Another particle B of mass '4 m' and charge 'q' is accelerated by a potential difference of $2500 V$ The ratio of de-Broglie wavelengths $\frac{λ_{A}}{λ_{B}}$ is close to :

10.00

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14.14

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4.47

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0.07

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Solution

The correct option is B $14.14$
K.E acquired by change = $K = q V$
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m K}} = \frac{h}{\sqrt{2 m q V}}$
$\frac{λ_{A}}{λ_{B}} = \frac{\sqrt{2 m_{B} q_{B} V_{B}}}{2 m_{A} q_{A} V_{A}} = 2 \sqrt{50}$
$= 2 \times 7.07 = 14.14$

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A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50V. Another particle B of mass '4 m' and charge 'q' is accelerated by a potential difference of 2500V The ratio of de-Broglie wavelengths λAλB is close to :

The correct option is B 14.14K.E acquired by change =K=qVλ=hp=h√2mK=h√2mqVλAλB=√2mBqBVB2mAqAVA=2√50=2×7.07=14.14

A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of $50 V$ . Another particle B of mass '4 m' and charge 'q' is accelerated by a potential difference of $2500 V$ The ratio of de-Broglie wavelengths $\frac{λ_{A}}{λ_{B}}$ is close to :

The correct option is B $14.14$
K.E acquired by change = $K = q V$
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m K}} = \frac{h}{\sqrt{2 m q V}}$
$\frac{λ_{A}}{λ_{B}} = \frac{\sqrt{2 m_{B} q_{B} V_{B}}}{2 m_{A} q_{A} V_{A}} = 2 \sqrt{50}$
$= 2 \times 7.07 = 14.14$