1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The de Broglie wavelength of electron of He+ ion is 3.32 Å. If the photon emitted upon de-excitation of this He+ ion is made to hit H-atom in its ground state so as to liberate electron from it, then the de Broglie wavelength of photoelectron is:

A
2.348 Å
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1.917 Å
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3.329 Å
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3.329 pm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 2.348 Å Energy of the photon liberated: Using formula E=hcλ Where h = planck’s constant 6.626×10−34 J s c = velocity of light = 3×108 m s−1 and λ is the wavelength (given) = 3.32 Å=3.32×10−10 m Putting all these values in the formula: E=6.626×10−34×3×1083.32×10−10 = 5.947×10−16J Now using formula for de-Broglie wavelength i.e., λ=h√m×E m = mass of electron = 9.109×10−31 kg λ=6.626×10−34√(9.109×10−31×5.947×10−16) = 6.626×10−345.4171×10−46 = 2.348 Å

Suggest Corrections
4
Join BYJU'S Learning Program
Related Videos
Fringe Width
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program