The wavelength of an electron of kinetic energy 4.50 x 10^-29 J is _________ x 10^-5 m. (Nearest integer) Given : mass of electron is 9 x 10^-31 kg, h =6.6 x 10^-34 J s

lambda_ d=frac h mv=frac hsqrt2 mKE=frac6.6 x 10^-34sqrt2 x 9 x 10^-31 x 4.5 x 10^-29 \ =frac6.6 x 10^-34sqrt9^2 x 10^-60 \ =frac6.6 x 10^-349 x 10^-30=(6.6 / 9) x 10^-4 \ =7.3 x 10^-5 m Therefore Ans =7

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Chemistry

Exam

Chemistry

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

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Correct Answer:

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

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