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

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Since the wavelengths are equal,we have $\frac{h}{m_p v_p} = \frac{h}{m_e v_e}$.Thi

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

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Since the wavelengths are equal,we have $\frac{h}{m_p v_p} = \frac{h}{m_e v_e}$.This simplifies to $m_p v_p = m_e v_e$.Given: mass of particle $m_p = 1 \, mg = 10^{-6} \, kg$,mass of electron $m_e = 9.1 	imes 10^{-31} \, kg$,and velocity of electron $v_e = 3 	imes 10^6 \, m/s$.Substituting the values: $10^{-6} \, kg 	imes v_p = 9.1 	imes 10^{-31} \, kg 	imes 3 	imes 10^6 \, m/s$.$v_p = \frac{27.3 	imes 10^{-25}}{10^{-6}} \, m/s = 27.3 	imes 10^{-19} \, m/s = 2.73 	imes 10^{-18} \, m/s$.Rounding to the nearest option,the velocity of the particle is $2.7 	imes 10^{-18} \, m/s$.

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

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