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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.Let the mass of particle $A$ be $m_A = m$ and the velocity be

Vedclass · Accepted Answer

$1 : 2$

Vedclass · Answer

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.Let the mass of particle $A$ be $m_A = m$ and the velocity be $v_A = 8v$.Then the mass of particle $B$ is $m_B = 4m$ and the velocity is $v_B = v$.The wavelength of particle $A$ is $\lambda_A = \frac{h}{m_A v_A} = \frac{h}{m 	imes 8v} = \frac{h}{8mv}$.The wavelength of particle $B$ is $\lambda_B = \frac{h}{m_B v_B} = \frac{h}{4m 	imes v} = \frac{h}{4mv}$.The ratio of the wavelengths is $\frac{\lambda_A}{\lambda_B} = \frac{h / 8mv}{h / 4mv} = \frac{4mv}{8mv} = \frac{1}{2}$.Thus,the ratio is $1 : 2$.

If the mass of particle $B$ is four times the mass of particle $A$ and the velocity of particle $A$ is eight times the velocity of particle $B$,then the ratio of the de Broglie wavelength of particle $A$ to particle $B$ will be .......

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