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

Vedclass · Accepted Answer

$1.05 	imes 10^{-34} \ J \cdot s$

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(C) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.The change in angular momentum $\Delta L$ when the electron moves from orbit $n_1 = 4$ to $n_2 = 5$ is:$\Delta L = L_2 - L_1 = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{h}{2\pi}(n_2 - n_1)$.Substituting the given values:$\Delta L = \frac{6.6 	imes 10^{-34}}{2 	imes 3.14} 	imes (5 - 4)$.$\Delta L = \frac{6.6 	imes 10^{-34}}{6.28} 	imes 1$.$\Delta L \approx 1.05 	imes 10^{-34} \ J \cdot s$.

When an electron in a hydrogen atom is excited from its $4^{th}$ to $5^{th}$ stationary orbit,the change in angular momentum of the electron is (Planck's constant: $h = 6.6 \times 10^{-34} \ J \cdot s$)

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