Question

) The life time of an electron in an excited state is
about 10-8 sec. The uncertainty in energy during
this time is:
(a) 10.5 x 10-19 J
(b) 10.5 x 10-29
(c) 1.05 x 10-16
(d) 1.05 x 10-16)​

Answer 1

Answer:

The uncertainty in energy during this time is equal to 5.27×10⁻²⁷J.

Explanation:

Given:

The life time of an electron in an excited state Δt = 10⁻⁸sec

Uncertainty in energy is given by ΔE,

From the uncertainty principle,

ΔE.Δt ≥ h/4π                                                   ...............(1)

where h is Plank's constant , $h=6.626*10^{-34}Js$

Put the value of  Δt and h in eq.(1);

ΔE(10⁻⁸s) ≈ 6.626×10⁻³⁴Js/4×3.14

ΔE  $=\frac{6.626*10^{-34}Js}{4*3.14*10^{-8}s}$

ΔE = $0.527*10^{-26}J$

ΔE = $5.27*10^{-27}J$

Therefore, the uncertainty in energy is $5.27*10^{-27}J$.

Answer 2

Answer:

The uncertainty in energy during this time is equal to 5 . 2 7 × 1 0 ⁻ ² ⁷ J.

Explanation:

Given : The life time of an electron in an excited state is

about 10 - 8 sec.

To find: The uncertainty in energy during

this time is ?

Solution:

During this period, the uncertainty in energy is equivalent to 5 . 2 7 1 0 2  7 J .

The life duration of an energised electron is t = 1 0 8 seconds.

E represents energy uncertainty.

According to the uncertainty principle,

Where h is Plank' s constant , E . t h / 4

In e q . ( 1 ) , enter the values of t and h;

Δ E ( 1 0 ⁻ ⁸ s ) = 6 . 6 2 6 × 1 0 ⁻ ³ ⁴ J s / 4 × 3 . 1 4

As a result, the energy uncertainty is.

5 . 2 7 × 1 0 ⁻ ² ⁷ J .

5 . 2 7 × 1 0 ⁻ ² ⁷ J . Is the correct answer of this question .

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