What is the de Broglie wavelength of a tennis | vedclass

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Question

(A) The de Broglie wavelength is given by the formula: $\lambda = \frac{h}{mv}$ Given: $h = 6.626 	imes 10^{-34} \ J \cdot s$ $m = 66 \ g = 66 	imes

Vedclass · Accepted Answer

$10^{-33} \ m$

Vedclass · Answer

(A) The de Broglie wavelength is given by the formula: $\lambda = \frac{h}{mv}$ Given: $h = 6.626 	imes 10^{-34} \ J \cdot s$ $m = 66 \ g = 66 	imes 10^{-3} \ kg = 6.6 	imes 10^{-2} \ kg$ $v = 10 \ m/s$ Substituting the values: $\lambda = \frac{6.626 	imes 10^{-34}}{6.6 	imes 10^{-2} 	imes 10} \ m$ $\lambda \approx \frac{6.626 	imes 10^{-34}}{6.6 	imes 10^{-1}} \ m$ $\lambda \approx 1.004 	imes 10^{-33} \ m$ Thus,the value is approximately $10^{-33} \ m$.

What is the de Broglie wavelength of a tennis ball weighing $66 \ g$ moving with a velocity of $10 \ m/s$?

Similar Questions

Calculate the wavelength of an electron if its mass is $9.1 \times 10^{-31} \, kg$,its velocity is $1/10$ of the speed of light,and the value of Planck's constant $h$ is $6.626 \times 10^{-34} \, J \cdot s$.

The de Broglie wavelength of an electron travelling with $20 \%$ of velocity of light is
$(h = 6.626 \times 10^{-34} \ J \ s; m_{e} = 9.1 \times 10^{-31} \ kg)$

The wavelength associated with the motion of an electron:

If the radius of the first orbit of hydrogen atom is $a_0$, then de Broglie's wavelength of electron in $3^{rd}$ orbit is (in $\pi a_0$)

Which of the following expressions represents the de-Broglie relationship?

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