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(C) In a hydrogen atom,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6/n^2 \ eV$. When an electron transitions from a higher

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Angular momentum of the electron remains constant

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(C) In a hydrogen atom,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6/n^2 \ eV$. When an electron transitions from a higher energy level to a lower energy level,$n$ decreases.As $n$ decreases,the radius of the orbit $r_n \propto n^2$ decreases.The kinetic energy $K = |E_n| = 13.6/n^2 \ eV$ increases as $n$ decreases.The velocity $v_n \propto 1/n$ increases as $n$ decreases.The angular momentum $L = mvr = n(h/2\pi)$ is quantized and depends on the principal quantum number $n$. Since $n$ changes during the transition,the angular momentum does not remain constant.The de-Broglie wavelength $\lambda = h/p = h/(mv)$. Since $v$ increases,$\lambda$ decreases.Therefore,the statement that the angular momentum remains constant is incorrect.

An electron in a hydrogen atom undergoes a transition from a higher energy level to a lower energy level. The incorrect statement of the following is

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