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(D) The magnetic force provides the necessary centripetal force for the circular motion of the electron: $e v_n B = \frac{m_e v_n^2}{r_n} \Rightarrow

Vedclass · Accepted Answer

$K.E = \frac{nhBe}{4\pi m_e}$

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(D) The magnetic force provides the necessary centripetal force for the circular motion of the electron: $e v_n B = \frac{m_e v_n^2}{r_n} \Rightarrow r_n = \frac{m_e v_n}{e B}$.According to Bohr's quantization postulate,the angular momentum is $L = m_e v_n r_n = \frac{nh}{2\pi}$.Substituting the expression for $r_n$ into the quantization condition:$m_e v_n \left( \frac{m_e v_n}{e B} \right) = \frac{nh}{2\pi} \Rightarrow m_e^2 v_n^2 = \frac{nh e B}{2\pi} \Rightarrow v_n^2 = \frac{nh e B}{2\pi m_e^2}$.The kinetic energy is $K.E = \frac{1}{2} m_e v_n^2$.Substituting $v_n^2$: $K.E = \frac{1}{2} m_e \left( \frac{nh e B}{2\pi m_e^2} \right) = \frac{nh e B}{4\pi m_e}$.

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