Question

In an isolated li2+ electron jumps from n=2 to n=1 .What is the recoil momentum of the ion

Answer 1

recoil momentum of the ion is 48.96 × 10^-27 Kgm/s

first find change in energy ,∆E = 13.6Z²(1/n1² - 1/n2²) eV

= 13.6 × (3)²[1 - 1/2²] eV

= 13.6 × 9 [1- 1/4] eV

= 13.6 × 9 × 3/4 eV

= 91.8 eV

= 91.8 × 1.6 × 10^-19 J

= 146.88 × 10^-19 J

= 1.4688 × 10^-17 J

now recoil momentum, P = ∆E/c

where c is speed of light in vaccum.

so, P = 1.4688 × 10^-17/(3 × 10^8)

= 0.4896 × 10^-25 Kgm/s

= 48.96 × 10^-27 Kgm/s

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Answer 2

$\huge\bold\purple{Answer:-}$

first find change in energy ,∆E = 13.6Z²(1/n1² - 1/n2²) eV

= 13.6 × (3)²[1 - 1/2²] eV

= 13.6 × 9 [1- 1/4] eV

= 13.6 × 9 × 3/4 eV

= 91.8 eV

= 91.8 × 1.6 × 10^-19 J

= 146.88 × 10^-19 J

= 1.4688 × 10^-17 J

now recoil momentum, P = ∆E/c

where c is speed of light in vaccum.

so, P = 1.4688 × 10^-17/(3 × 10^8)

= 0.4896 × 10^-25 Kgm/s

= 48.96 × 10^-27 Kgm/s