Question

CHEMISTRY : IIT-JEE/NEET-OLYMPIADS
IX-ATOMIC STRUCTURE
13. Calculate the momentum of a particle which has a de-Broglie wavelength of 2.5x10-1°m.
(h=6.625x10-34 kg m’s!)
1) 5.32x10-"Kgm sec
2) 2.65 x10 24 Kgm sec
4) 9.2x10-20Kgm sec
3) 1.38x10-'Kg m sec
cubitalis​

Answer 1

Answer:

option 2 is somewhat similar to the actual answer

Step-by-step explanation:

we have to find MV i.e. momentum

the relation between the de Broglie wavelength and momentum of the particle is given by

L = h /MV , where h is Planck's constant whose value is 6.625 × 10 ^-34

so

MV = 6.625 × 10^ -34 / 2.5 ×10^-1

MV = 2.65 × 10^-33 kg m/s