KCET · Chemistry · Structure of Atom
If wavelength of photon is \(2.2 \times 10^{-11} \mathrm{~m}\) and \(h=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\), then momentum of photon
- A \(3.33 \times 10^{-22} \mathrm{~kg} \mathrm{~ms}^{-1}\)
- B \(1.452 \times 10^{-44} \mathrm{~kg} \mathrm{~ms}^{-1}\)
- C \(6.89 \times 10^{+43} \mathrm{~kg} \mathrm{~ms}^{-1}\)
- D \(3 \times 10^{-23} \mathrm{~kg} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(3 \times 10^{-23} \mathrm{~kg} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, wavelength of photon, \(\lambda=22 \times 10^{-11} \mathrm{~m}\)
Plank constant, \(h=6.6 \times 10^{-34} \mathrm{Js}\)
We know that, \(\lambda=\frac{h}{m c} \Rightarrow m c=\frac{h}{\lambda}\) and \(m c=\) momentum
\[
\begin{aligned}
\therefore \text { Momentum } & =\frac{6.6 \times 10^{-34}}{2.2 \times 10^{-11}} \\
& =3 \times 10^{-23} \mathrm{~kg} \mathrm{~ms}^{-1}
\end{aligned}
\]
Plank constant, \(h=6.6 \times 10^{-34} \mathrm{Js}\)
We know that, \(\lambda=\frac{h}{m c} \Rightarrow m c=\frac{h}{\lambda}\) and \(m c=\) momentum
\[
\begin{aligned}
\therefore \text { Momentum } & =\frac{6.6 \times 10^{-34}}{2.2 \times 10^{-11}} \\
& =3 \times 10^{-23} \mathrm{~kg} \mathrm{~ms}^{-1}
\end{aligned}
\]
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