Draw the curve showing the variation of de Broglie wavelength of a particle with its momentum.
Find the momentum of a photon of wavelength `0.01 Å`

Explanation

1st part: If momentum = p, then the de Broglie wavelength of a particle,
`lamda = (h)/(p)`
or, `lamdap = h`= Planck's constant Hence the `lamda - p` graph will be a rectangular hyperbola, as shown in the figure.

Given, `lamda = 0.01 Å = 10^(-12)m`
2nd part: Momentum of the photon,
`p = (E)/(c) = (hf)/(c) = (h)/(lamda) = (6.63 xx 10^(-34))/(10^(-12))`
`= 6.63 xx 10^(-22) kg.m.s^(-1)`