An atom is exated Date which, has the energy of 3 ev when it makes transition to ground state the Wavelength of the a light emitted is
Answers
Answer:
Correct option is
B
E=3.4eV,λ=6.6×10
−10
m
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=−2×3.4+KE
KE=3.4eV
Let p be the momentum of electron and m be the mass of electron.
E=
2m
p
2
p=
2
mE
Now, the De-Broglie wavelength associated with an electron is:
λ=
p
h
λ=
2
mE
h
λ=
2
×9.1×10
−31
×(−3.4)×1.6×10
−19
6.6×10
34
λ=
9.95×10
−25
6.6×10
34
λ=0.66×10
−9
λ=6.6×10
−10
m
Answer:
B
E=3.4 eV, λ=−6.6×10
−10
m
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=
2m
p
2
p=
2
mE
Now, the De-Broglie wavelength associated with an electron is
λ=
p
h
λ=
2
mE
h
λ=
2
×9.1×10
−31
×(−3.4)×1.6×10
−19
6.6×10
−34
λ=−
9.95×10
−25
6.6×10
−34
λ=−0.66×10
−9
λ=−6.6×10
−10
m
Explanation:
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