If the kinetic energy of an electron is 2.8 t | vedclass

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Question

(C) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.Given: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$m = 9

Vedclass · Accepted Answer

$9.28 	imes 10^{-8} \ m$

Vedclass · Answer

(C) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.Given: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$m = 9.1 	imes 10^{-31} \ kg$,$E = 2.8 	imes 10^{-23} \ J$.Substituting the values:$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 2.8 	imes 10^{-23}}}$$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{50.96 	imes 10^{-54}}}$$\lambda = \frac{6.63 	imes 10^{-34}}{7.138 	imes 10^{-27}}$$\lambda \approx 9.28 	imes 10^{-8} \ m$.

If the kinetic energy of an electron is $2.8 \times 10^{-23} \ J$,then the de Broglie wavelength will be ....... $(m_e = 9.1 \times 10^{-31} \ kg)$.

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