Question

What is the wavelength associated with a particle of mass 0.1 moving with a speed of 1 x 105cm Sec (h = 6.6 x 10^27 erg Sec) Answer, 6.6 x 20^-37cm

Answer 1

Combining the following equations,

E = mc² and E = hc/λ

we get,

λ = h/mv

So, λ = 6.6 * 10⁻²⁷ / (0.1 * 10⁵)

λ = 6.6 * 10⁻³¹ cm

Answer 2

Answer: The wavelength associated with the particle is $6.6 * 10^{-31} cm$.

Explanation:

Given: Mass of particle, $m = 0.1 g$

Velocity of particle ,$v = 1 * 10^{5} cm s^{-1}$

Plank's constant,  $h = 6.6 * 10^{-27} erg sec$

To find: wavelength of the particle, λ

Solution: The De-Broglie equation relates the wavelength (λ) and momentum (mv) of a particle as:

λ $=\frac{h}{mv}$

Putting the values in the above equation:

λ  $= \frac{6.6 * 10^{-27}erg }{0.1 g * 1 * 10^{5} cm s^{-1} } = \frac{6.6 * 10^{-27} g cm^{2}s^{-1} }{0.1 g * 1 * 10^{5} cm s^{-1} }= 6.6 * 10^{-31} cm$

(As $1 erg = 1 gcm^{2} s^{-1}$)

Therefore the wavelength associated with the particle is $6.6 * 10^{-31} cm$.

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