A hydrogen atom in its ground state absorbs 1 | vedclass

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(B) The energy absorbed by the hydrogen atom is given by $\Delta E = E_n - E_1 = 13.6 \left( 1 - \frac{1}{n^2} \right) \; eV$. Given $\Delta E = 10.2

Vedclass · Accepted Answer

$1.05$

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(B) The energy absorbed by the hydrogen atom is given by $\Delta E = E_n - E_1 = 13.6 \left( 1 - \frac{1}{n^2} ight) \; eV$. Given $\Delta E = 10.2 \; eV$,we have $13.6 \left( 1 - \frac{1}{n^2} ight) = 10.2$. $1 - \frac{1}{n^2} = \frac{10.2}{13.6} = 0.75 = \frac{3}{4}$. $\frac{1}{n^2} = 1 - 0.75 = 0.25 = \frac{1}{4}$,so $n = 2$. The angular momentum of an electron in the $n$-th orbit is given by $L = \frac{nh}{2\pi}$. Initial angular momentum $(n=1)$: $L_i = \frac{1 \cdot h}{2\pi}$. Final angular momentum $(n=2)$: $L_f = \frac{2 \cdot h}{2\pi}$. The increase in angular momentum is $\Delta L = L_f - L_i = \frac{2h}{2\pi} - \frac{h}{2\pi} = \frac{h}{2\pi}$. Substituting $h = 6.6 	imes 10^{-34} \; J \cdot s$ and $\pi \approx 3.14$: $\Delta L = \frac{6.6 	imes 10^{-34}}{2 	imes 3.14} \approx 1.05 	imes 10^{-34} \; J \cdot s$.

$A$ hydrogen atom in its ground state absorbs $10.2 \; eV$ of energy. The angular momentum of the electron of the hydrogen atom will increase by the value of ............... $\times 10^{-34} \; J \cdot s$ (Given,Planck's constant $h = 6.6 \times 10^{-34} \; J \cdot s$)

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