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Question

(C) The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{1^2} - \frac{1}{5^2} \right]

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$\frac{24hR}{25m}$

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(C) The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{1^2} - \frac{1}{5^2} \right] = R \left[ 1 - \frac{1}{25} \right] = \frac{24R}{25}$.By the conservation of momentum,the momentum of the atom $P_{atom}$ must be equal to the momentum of the emitted photon $P_{photon}$.$P_{photon} = \frac{h}{\lambda}$.Since the atom was initially stationary,$m_{atom} v = \frac{h}{\lambda}$,where $m$ is the mass of the hydrogen atom.Substituting $\frac{1}{\lambda} = \frac{24R}{25}$ into the equation:$v = \frac{h}{m} \cdot \frac{1}{\lambda} = \frac{h}{m} \cdot \frac{24R}{25} = \frac{24hR}{25m}$.

$1$. Burning candle	$i$. Line spectrum
$2$. Sodium vapour	$ii$. Continuous spectrum
$3$. Bunsen flame	$iii$. Band spectrum
$4$. Dark lines in solar spectrum	$iv$. Absorption spectrum

List-$I$	List-$II$
$1$. Nitrogen molecules	$(A)$ Continuous spectrum
$2$. Incandescent solids	$(B)$ Absorption spectrum
$3$. Fraunhofer lines	$(C)$ Band spectrum
$4$. Electric arc between iron rods	$(D)$ Emission spectrum

An electron of a stationary hydrogen atom transitions from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission will be:

Similar Questions

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$2$. Sodium vapour $ii$. Continuous spectrum
$3$. Bunsen flame $iii$. Band spectrum
$4$. Dark lines in solar spectrum $iv$. Absorption spectrum

The energy required to remove one electron from a helium atom is $24.6 \, eV$. What is the energy required to remove both electrons from the helium atom (in $, eV$)?

What is the atomic number of gold?

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$1$. Nitrogen molecules $(A)$ Continuous spectrum
$2$. Incandescent solids $(B)$ Absorption spectrum
$3$. Fraunhofer lines $(C)$ Band spectrum
$4$. Electric arc between iron rods $(D)$ Emission spectrum

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