When an electron is excited from its 4^{text{ | vedclass

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(C) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{	ext{th}}$ orbit is given by $L = \frac{nh}{2\pi}$.Change in ang

Vedclass · Accepted Answer

$1 	imes 10^{-34} 	ext{ J s}$

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(C) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{	ext{th}}$ orbit is given by $L = \frac{nh}{2\pi}$.Change in angular momentum $\Delta L = L_2 - L_1$.Here,$n_1 = 4$ and $n_2 = 5$.$\Delta L = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{h}{2\pi} (n_2 - n_1)$.Substituting the values: $\Delta L = \frac{6.63 	imes 10^{-34}}{2 	imes 3.14} (5 - 4)$.$\Delta L = \frac{6.63 	imes 10^{-34}}{6.28} \approx 1.055 	imes 10^{-34} 	ext{ J s}$.Rounding to the nearest given option,the change is approximately $1 	imes 10^{-34} 	ext{ J s}$.

When an electron is excited from its $4^{\text{th}}$ orbit to its $5^{\text{th}}$ stationary orbit,the change in the angular momentum of the electron is approximately (Planck's constant $= h = 6.63 \times 10^{-34} \text{ J s}$)

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