A particle of mass 1, mg has the same wavelen | vedclass

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Since the particle and the electron have the same wavelength,their momenta must be

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$2.7 	imes 10^{-18}\, m\,s^{-1}$

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Since the particle and the electron have the same wavelength,their momenta must be equal.$m_p v_p = m_e v_e$Here,$m_p = 1\, mg = 10^{-6}\, kg$,$m_e = 9.1 	imes 10^{-31}\, kg$,and $v_e = 3 	imes 10^6\, m\,s^{-1}$.Substituting the values:$v_p = \frac{m_e v_e}{m_p} = \frac{9.1 	imes 10^{-31} 	imes 3 	imes 10^6}{10^{-6}}$$v_p = 27.3 	imes 10^{-25} 	imes 10^6 = 27.3 	imes 10^{-19} = 2.73 	imes 10^{-18}\, m\,s^{-1}$.Rounding to the given options,the velocity is $2.7 	imes 10^{-18}\, m\,s^{-1}$.

$A$ particle of mass $1\, mg$ has the same wavelength as an electron moving with a velocity of $3 \times 10^6\, m\,s^{-1}$. The velocity of the particle is (Mass of electron $= 9.1 \times 10^{-31}\, kg$)

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