Question

CHEMISTRY

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 1.5×107ms−1,  calculate the energy with which it is bound to the nucleus. chemisrty class 11 cbse question.​

Answer 1

Given :-

• Wavelength of photon, λ = 150 pm = 150 × 10⁻¹² m.
• Velocity of ejected electron, v = 1.5 × 10⁷ ms⁻¹.

To Find :-

• Energy with which the electron was bound to the nucleus.

Solution :-

We know, energy of incident photon = $\rm{\dfrac{hc}{\lambda}}$, where:-

▪Planck's constant, h = 6.626 × 10⁻³⁴ Js.

▪Speed of light, c = 3.0 × 10⁸ ms⁻¹.

Putting the values, we get,

$\implies\rm{\dfrac{(6.626\times{}10^{-34}\:Js)(3.0\times{}10^8\:ms^{-1})}{(150\times{}10^{-12}\:m)}}$

$\implies\rm{13.25\times{}10^{-16}\:J}$

Now, Energy of the ejected electron = $\rm{\dfrac{1}{2}mv^2}$ where:-

▪Mass of electron, m = 9.11 × 10⁻³¹ kg.

▪Velocity of electron, v = 1.5 × 10⁷ ms⁻¹.

Putting the values, we get,

$\implies\rm{\dfrac{1}{2}\times{}9.11\times{}10^{-31}\:kg\times{}(1.5\times{}10^7\:ms^{-1})^2}$

$\implies\rm{1.025\times{}10^{-16}\:J}$

$\rule{350}{3}$

Energy with which electron was bound to the nucleus =

⇒ Energy of incident photon - energy of electron

$\implies\rm{13.25\times{}10^{-16}\:J-1.025\times{}10^{-16}\:J}$

$\implies\rm{12.225\times{}10^{-16}\:J=}\:\bold{7.63\times{}10^3\:eV.}$