Question

Calculate the value of the uncertainty in the energy of the excited state of an atom when the electron is lived in excited state for 8.4*10^-3 s
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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

Answer 2

we have to calculate the value of the uncertainty in the energy of the excited state of an atom when the electron is lived in excited state for 8.4 × 10¯³ sec.

solution : using Heisenberg's uncertainty principle, ∆E × ∆t = h/4π

here ∆t = 8.4 × 10¯³ sec

h = 6.63 × 10¯³⁴ J.s

so, E = h/4π∆t

= (6.63 × 10¯³⁴ J.s)/(4 × 3.14 × 8.4 × 10¯³ s)

= 0.063 × 10¯³¹ J

= 6.3 × 10¯³³ J

Therefore the uncertainty in the energy of excited state of an atom is 6.3 × 10¯³³ J

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