JEE MAIN - Chemistry (2022 - 28th July Evening Shift - No. 15)
If the wavelength for an electron emitted from $$\mathrm{H}$$-atom is $$3.3 \times 10^{-10} \mathrm{~m}$$, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is _________ times. (Nearest integer)
$$\left[\right.$$ Given $$: \mathrm{h}=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}$$ ]
Mass of electron $$=9.1 \times 10^{-31} \mathrm{~kg}$$
Answer
2
Explanation
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$
$$ \begin{aligned} &\Rightarrow \mathrm{mv}=\frac{\mathrm{h}}{\lambda}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \frac{\mathrm{m}^{2}}{\mathrm{sec}^{2}} \times \mathrm{sec}}{3.3 \times 10^{-10} \mathrm{~m}} \\\\ &\mathrm{mv}=\frac{6.626 \times 10^{-24}}{3.3}=2 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \,\mathrm{sec}{ }^{-1} \end{aligned} $$
Kinetic energy $=\frac{1}{2} m v^{2}$
$$ \begin{aligned} &=\frac{(\mathrm{mv})^{2}}{2 \mathrm{~m}} \\\\ &=\frac{\left(2 \times 10^{-24}\right)^{2}}{2 \times 9.1 \times 10^{-31} \mathrm{~kg}} \\\\ &=2.18 \times 10^{-18} \mathrm{~J} \\\\ &=21.8 \times 10^{-19} \mathrm{~J} \end{aligned} $$
Total energy absorbed $=$ lonization energy $+$ Kinetic energy
$$ \begin{aligned} &=(21.76+21.8) \times 10^{-19} \\\\ &=43.56 \times 10^{-19} \mathrm{~J} \\\\ &\approx 2 \text { times of } 21.76 \times 10^{-19} \mathrm{~J} \end{aligned} $$
$$ \begin{aligned} &\Rightarrow \mathrm{mv}=\frac{\mathrm{h}}{\lambda}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \frac{\mathrm{m}^{2}}{\mathrm{sec}^{2}} \times \mathrm{sec}}{3.3 \times 10^{-10} \mathrm{~m}} \\\\ &\mathrm{mv}=\frac{6.626 \times 10^{-24}}{3.3}=2 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \,\mathrm{sec}{ }^{-1} \end{aligned} $$
Kinetic energy $=\frac{1}{2} m v^{2}$
$$ \begin{aligned} &=\frac{(\mathrm{mv})^{2}}{2 \mathrm{~m}} \\\\ &=\frac{\left(2 \times 10^{-24}\right)^{2}}{2 \times 9.1 \times 10^{-31} \mathrm{~kg}} \\\\ &=2.18 \times 10^{-18} \mathrm{~J} \\\\ &=21.8 \times 10^{-19} \mathrm{~J} \end{aligned} $$
Total energy absorbed $=$ lonization energy $+$ Kinetic energy
$$ \begin{aligned} &=(21.76+21.8) \times 10^{-19} \\\\ &=43.56 \times 10^{-19} \mathrm{~J} \\\\ &\approx 2 \text { times of } 21.76 \times 10^{-19} \mathrm{~J} \end{aligned} $$
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