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If the photon of the wavelength \(150 \mathrm{pm}\) strikes an atom and one of its inner bound electrons is ejected out with a velocity of \(1.5 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}\), calculate the energy with which it is bound to the nucleus.
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Verified Answer
Energy of the incident photon
\(\begin{aligned}
&=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^8 \mathrm{~ms}^{-1}\right)}{\left(150 \times 10^{-12} \mathrm{~m}\right)} \\
&=13.25 \times 10^{-16} \mathrm{~J}
\end{aligned}\)
Energy of the electron ejected \(=\frac{1}{2} \mathrm{mv}^2\)
\(\begin{aligned}
&=\frac{1}{2}\left(9.11 \times 10^{-31} \mathrm{~kg}\right)\left(1.5 \times 10^7 \mathrm{~ms}^{-1}\right)^2 \\
&=1.025 \times 10^{-16} \mathrm{~J}
\end{aligned}\)
Energy with which the electron was bound to the nucleus \(=13.25 \times 10^{-16} \mathrm{~J}-1.025 \times 10^{-16} \mathrm{~J}=12.225 \times 10^{-16} \mathrm{~J}\)
\(=\frac{12.225 \times 10^{-16}}{1.602 \times 10^{-19}} \mathrm{eV}=7.63 \times 10^3 \mathrm{eV} \text {. }\)
\(\begin{aligned}
&=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^8 \mathrm{~ms}^{-1}\right)}{\left(150 \times 10^{-12} \mathrm{~m}\right)} \\
&=13.25 \times 10^{-16} \mathrm{~J}
\end{aligned}\)
Energy of the electron ejected \(=\frac{1}{2} \mathrm{mv}^2\)
\(\begin{aligned}
&=\frac{1}{2}\left(9.11 \times 10^{-31} \mathrm{~kg}\right)\left(1.5 \times 10^7 \mathrm{~ms}^{-1}\right)^2 \\
&=1.025 \times 10^{-16} \mathrm{~J}
\end{aligned}\)
Energy with which the electron was bound to the nucleus \(=13.25 \times 10^{-16} \mathrm{~J}-1.025 \times 10^{-16} \mathrm{~J}=12.225 \times 10^{-16} \mathrm{~J}\)
\(=\frac{12.225 \times 10^{-16}}{1.602 \times 10^{-19}} \mathrm{eV}=7.63 \times 10^3 \mathrm{eV} \text {. }\)
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