Question 464: Atoms & Nuclei

Question

An electron of stationary Hydrogen atom passes from fifth energy level to ground level. The velocity that the atom acquired as a result of photo emission is
( $\mathrm{m}=$ mass of electron, $\mathrm{R}=$ Rydberg's constant)
( $\mathrm{h}=$ Planck's constant)

Accepted Answer

$$\begin{aligned}
&\frac{1}{\lambda}=R\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2{ }^2}ight)\
&	ext { In this case, } \mathrm{n}_1=1 	ext { and } \mathrm{n}_2=5\
&	herefore \quad \frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{5^2}ight)=\frac{24}{25} R\quad	ext{... (i)}
\end{aligned}$$
Momentum of Photon,
$$p=\frac{h}{\lambda}=h\left(\frac{24}{25} Right)\quad 	ext{...[From (i)]
}$$
By conservation of momentum,
Momentum of Photon = Momentum of atom
$$\begin{aligned}
& h\left(\frac{24}{25} Right)=m v \
	herefore \quad & v=\frac{24 R h}{25 m}
\end{aligned}$$

Answer

$\frac{24 \mathrm{Rh}}{25 \mathrm{~m}}$

Answer

$\frac{25 \mathrm{Rh}}{24 \mathrm{~m}}$

Answer

$\frac{25 \mathrm{~m}}{24 \mathrm{Rh}}$

Answer

$\frac{24 \mathrm{~m}}{25 \mathrm{Rh}}$

Atoms & Nuclei

An electron of stationary Hydrogen atom passes from fifth energy level to ground level. The velocity that the atom acquired as a result of photo emission is ( \mathrm{m}= mass of electron, \mathrm{R}= Rydberg's constant) ( \mathrm{h}= Planck's constant)

Choose the correct answer: