A photon of 19.6 eV energy strikes a H-atom (in its ground state). Find de-Broglie wavelength
of electron ejected from H-atom (in Å).
Answers
Answered by
3
Answer is 3.32 
Explanation:
De Broglie wavelength () is given by the equation
- h=Planck’s constant whose value is
joule-seconds and
- p = momentum of the particle(here electron)
In terms of kinetic energy(E) momentum(p) can be written as,
where m=mass of the particle.
Hence becomes
"""(1)
Given here,
- E =
joule
- m(mass of electron)=
kg
Putting these values in equation (1) we get ,
λ = meter
= meter
=
Answered by
5
Answer:
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