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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)
(mass of electron $= 9.1 \times 10^{-31}$ kg)
A
$3 \times 10^{-31}$ m s$^{-1}$
B
$2.7 \times 10^{-21}$ m s$^{-1}$
C
$2.7 \times 10^{-18}$ m s$^{-1}$
D
$9 \times 10^{-2}$ m s$^{-1}$
Detailed Solution
de Broglie wavelength associated with the electron moving with velocity $v_e$: $\lambda_e = \dfrac{h}{m_e v_e}$
$\lambda_e = \dfrac{h}{9.1 \times 10^{-31} \times 3 \times 10^{6}}$
Wavelength of the particle of mass 1 mg $= 1 \times 10^{-6}$ kg moving with velocity $v$: $\lambda_p = \dfrac{h}{1 \times 10^{-6} \times v}$
As given, $\lambda_e = \lambda_p$
$\Rightarrow \dfrac{h}{9.1 \times 10^{-31} \times 3 \times 10^{6}} = \dfrac{h}{10^{-6} \times v}$
$\Rightarrow 10^{-6} \times v = 9.1 \times 10^{-31} \times 3 \times 10^{6}$
$\Rightarrow v = \dfrac{27.3 \times 10^{-25}}{10^{-6}}$ m/s
$v = 2.73 \times 10^{-18}$ m/s $\approx 2.7 \times 10^{-18}$ m s$^{-1}$
$\lambda_e = \dfrac{h}{9.1 \times 10^{-31} \times 3 \times 10^{6}}$
Wavelength of the particle of mass 1 mg $= 1 \times 10^{-6}$ kg moving with velocity $v$: $\lambda_p = \dfrac{h}{1 \times 10^{-6} \times v}$
As given, $\lambda_e = \lambda_p$
$\Rightarrow \dfrac{h}{9.1 \times 10^{-31} \times 3 \times 10^{6}} = \dfrac{h}{10^{-6} \times v}$
$\Rightarrow 10^{-6} \times v = 9.1 \times 10^{-31} \times 3 \times 10^{6}$
$\Rightarrow v = \dfrac{27.3 \times 10^{-25}}{10^{-6}}$ m/s
$v = 2.73 \times 10^{-18}$ m/s $\approx 2.7 \times 10^{-18}$ m s$^{-1}$