Respuesta :
According to de Broglie, the wavelength of an object is related to its mass (m) and velocity (v) by the following equation:
\lambda = h/m*v ----------(1)
where h = planck's constant = 6.626 * 10^-34 m2 kg/s
Mass of the bullet (m) = 5.01 g = 5.01 * 10^-3 kg
Velocity v = 1097 miles/hr = 1097 * 1609.34 m/3600s = 490.40 m/s
Based on equation(1)
wavelength = 6.626*10^34 m2kgs-1 / 5.01*10^-3 kg*490.40 ms-1
= 2.69 * 10^34 m
Answer:
Wavelength of the bullet, [tex]\lambda=2.69\times 10^{-34}\ m[/tex]
Explanation:
It is given that,
Mass of the bullet, m = 5.01 g = 0.00501 kg
Velocity of the bullet, v = 1097 miles per hour = 490.4029 m/s
We need to find the de Broglie wavelength of the bullet. The de Broglie wavelength of the bullet is given by the following formula as :
[tex]\lambda=\dfrac{h}{mv}[/tex]
[tex]\lambda=\dfrac{6.62\times 10^{-34}\ J.s}{0.00501\ kg\times 490.4029\ m/s}[/tex]
[tex]\lambda=2.69\times 10^{-34}\ m[/tex]
So, the wavelength of the bullet is [tex]2.69\times 10^{-34}\ m[/tex]. Hence, this is the required solution.
