MHT CET · Physics · Center of Mass Momentum and Collision
An electron moving with velocity \(1.6 \times 10^7 \mathrm{~m} / \mathrm{s}\) has wavelength of \(0.4 Å\). The required accelerating voltage for the electron motion is [charge on electron \(=1.6 \times 10^{-19} \mathrm{C}\), mass of electron \(\left.=9 \times 10^{-31} \mathrm{~kg}\right]\)
- A \(7.2 \times 10^3 \mathrm{~V}\)
- B \(7.2 \times 10^2 \mathrm{~V}\)
- C \(7.2 \mathrm{~V}\)
- D \(7.2 \times 10^{-2} \mathrm{~V}\)
Answer & Solution
Correct Answer
(B) \(7.2 \times 10^2 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
When an electron is accelerated through a voltage its kinetic energy is converted into electric potential energy:
\(\mathrm{K} =\mathrm{U} \)
\( \frac{1}{2} \mathrm{mv}^2=\mathrm{eV} \)
\( \therefore \mathrm{V} =\frac{\mathrm{mv}^2}{2 \mathrm{e}} \)
\( \mathrm{V}=\frac{\left(9 \times 10^{-31}\right)\left(1.6 \times 10^7\right)^2}{2 \times 1.6 \times 10^{-19}}=7.2 \times 10^2 \mathrm{~V}\)
\(\mathrm{K} =\mathrm{U} \)
\( \frac{1}{2} \mathrm{mv}^2=\mathrm{eV} \)
\( \therefore \mathrm{V} =\frac{\mathrm{mv}^2}{2 \mathrm{e}} \)
\( \mathrm{V}=\frac{\left(9 \times 10^{-31}\right)\left(1.6 \times 10^7\right)^2}{2 \times 1.6 \times 10^{-19}}=7.2 \times 10^2 \mathrm{~V}\)
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