JEE MAIN - Chemistry (2023 - 6th April Morning Shift - No. 15)
The wavelength of an electron of kinetic energy $$4.50\times10^{-29}$$ J is _________ $$\times 10^{-5}$$ m. (Nearest integer)
Given : mass of electron is $$9\times10^{-31}$$ kg, h $$=6.6\times10^{-34}$$ J s
Answer
7
Explanation
$$
\begin{aligned}
& \lambda_{\mathrm{d}}=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 4.5 \times 10^{-29}}} \\\\
& =\frac{6.6 \times 10^{-34}}{\sqrt{9^2 \times 10^{-60}}} \\\\
& =\frac{6.6 \times 10^{-34}}{9 \times 10^{-30}}=\frac{6.6}{9} \times 10^{-4} \\\\
& =7.3 \times 10^{-5} \mathrm{~m}
\end{aligned}
$$
Therefore Ans $=7$
Therefore Ans $=7$
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