The wavelength of an electron in the first or | vedclass

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(C) From de-Broglie's equation: $p = \frac{h}{\lambda}$ or $v = \frac{h}{m \lambda}$ Kinetic Energy ($K$.$E$.) $= \frac{1}{2} mv^2 = \frac{1}{2} m \le

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$2.22 	imes 10^{-18}$

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(C) From de-Broglie's equation: $p = \frac{h}{\lambda}$ or $v = \frac{h}{m \lambda}$ Kinetic Energy ($K$.$E$.) $= \frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{h}{m \lambda}ight)^2 = \frac{h^2}{2m \lambda^2}$ Substituting the given values: $K$.$E$. $= \frac{(6.6 	imes 10^{-34})^2}{2 	imes 9.0 	imes 10^{-31} 	imes (3.3 	imes 10^{-10})^2}$ $K$.$E$. $= \frac{43.56 	imes 10^{-68}}{18.0 	imes 10^{-31} 	imes 10.89 	imes 10^{-20}}$ $K$.$E$. $= \frac{43.56 	imes 10^{-68}}{196.02 	imes 10^{-51}} \approx 0.222 	imes 10^{-17} \ J$ $K$.$E$. $= 2.22 	imes 10^{-18} \ J$

The wavelength of an electron in the first orbit of a hydrogen atom is $3.3 \times 10^{-10} \ m$. The kinetic energy of the electron (in $J$) is:
$(h = 6.6 \times 10^{-34} \ Js, m_{e} = 9.0 \times 10^{-31} \ kg)$

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The wavelength of an electron in the first orbit of a hydrogen atom is $3.3 \times 10^{-10} \ m$. The kinetic energy of the electron (in $J$) is: $(h = 6.6 \times 10^{-34} \ Js, m_{e} = 9.0 \times 10^{-31} \ kg)$