If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 ms–1, - Chemistry

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Numerical

If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 ms–1, calculate the energy with which it is bound to the nucleus.

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Solution 1

Energy of incident photon (E) is given by,

`E = ("hc")/lambda`

`= ((6.626xx10^(-34) " Js")(3.0 xx 10^8 " ms"^(-1)))/(150xx10^(-12) " m")`

`= 1.3252 xx 10^(-15)` J

`= 13.252 xx 20^(-16)` J

Energy of the electron ejected (K.E)

`= 1/2 "m"_"e""v"^2`

`=1/2(9.10939xx10^(-31) " kg")(1.5 xx 10^7 " ms"^(-1))^2`

= 10.2480 × 10–17 J

= 1.025 × 10–16 J

Hence, the energy with which the electron is bound to the nucleus can be obtained as:

= E – K.E

= 13.252 × 10–16 J – 1.025 × 10–16 J

= 12.227 × 10–16 J

`= (12.227xx10^(-16))/(1.602xx10^(-19))` eV

`= 7.6 xx 10^3` eV

`(5lambda_0 - 2000)/(4lambda_0 - 20000) = (5.35/2.55)^2 = 28.6225/6.5025`

`(5lambda_0 - 2000)/(4lambda_0- 2000) = 4.40177`

`17.6070lambda_0 - 5lambda_0 = 8803.537- 2000`

`lambda_0 = (6805.537)/12.607`

`lambda_0 = 539.8 "nm"`

`lamda_0 = 540 "nm"`

Solution 2

Energy of the incident photon= hc/λ = (6.626×10-34 Js×3.0×10ms-1)/(150×10-12m) = 13.25×10-16 J

Energy of the electron ejected = 1/2 mv= 1/2×(9.11×10-31kg)×(1.5×107ms-1)= 1.025×10-16 J

Energy with which the electron was bound to the nucleus = 13.25×10-16 J - 1.025×10-16 J

= 12.225×10-16 J = 12.225×10-16/1.602×10-19 eV = 7.63×10eV

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Chapter 2: Structure of Atom - EXERCISES [Page 72]

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NCERT Chemistry Part 1 and 2 Class 11
Chapter 2 Structure of Atom
EXERCISES | Q 2.54 | Page 72

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