The frequency of the de-Broglie wave of an el | vedclass

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.The frequency $
u$ is given by $
u = \frac{v}{\lambda}$.Substituting $\lambda = \

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$661$

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.The frequency $
u$ is given by $
u = \frac{v}{\lambda}$.Substituting $\lambda = \frac{h}{mv}$,we get $
u = \frac{v \cdot mv}{h} = \frac{mv^2}{h}$.From Bohr's model,the kinetic energy $K.E. = \frac{1}{2}mv^2 = 2.18 	imes 10^{-18} \ J$.Therefore,$mv^2 = 2 	imes 2.18 	imes 10^{-18} = 4.36 	imes 10^{-18} \ J$.Now,calculate the frequency: $
u = \frac{4.36 	imes 10^{-18}}{6.6 	imes 10^{-34}} \ Hz$.$
u = 0.6606 	imes 10^{16} \ Hz = 660.6 	imes 10^{13} \ Hz$.Rounding to the nearest integer,we get $661 	imes 10^{13} \ Hz$.

The frequency of the de-Broglie wave of an electron in Bohr's first orbit of a hydrogen atom is . . . . . . . . . . $\times 10^{13} \ Hz$ (nearest integer).
[Given: $R_H$ (Rydberg constant) $= 2.18 \times 10^{-18} \ J$,$h$ (Planck's constant) $= 6.6 \times 10^{-34} \ J \cdot s$]

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The frequency of the de-Broglie wave of an electron in Bohr's first orbit of a hydrogen atom is . . . . . . . . . . $\times 10^{13} \ Hz$ (nearest integer).[Given: $R_H$ (Rydberg constant) $= 2.18 \times 10^{-18} \ J$,$h$ (Planck's constant) $= 6.6 \times 10^{-34} \ J \cdot s$]