Question

An electron of hydrogen atom has been excited to a level where its ke is 1.51 ev what is the wavelength

Answer 1

Answer:

The potential energy of an electron is twice the kinetic energy(with negative sign) of an electron.

PE=−2E

The total energy is:

TE=PE+KE

−3.4=−2E+E

E=3.4eV

Let, p be the momentum of electron and m be the mass of electron.

E=2mp2

p=2mE

Now, the De-Broglie wavelength associated with an electron is

λ=ph

λ=2mEh

λ=2×9.1×10−31×(−3.4)×1.6×10−196.6×10−34

λ=−9.95×10−256.6×10−34

λ=−0.66×10−9

λ=−6.6×10−10m

Explanation:

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