Question

Light of wavelength 2.40 pm is directed onto a target containing free electrons. (a) Find the wavelength of light scattered at 30.0⁰ from the incident direction. (b) Do the same for a scattering angle of 120⁰?

Short answer

a. The wavelength of light scattered at 30.0⁰ from the incident direction is 2.73 pm.

b. The wavelength of light scattered at 120.0⁰ from the incident direction is 6.05 pm

Step-by-step solution

Identification of the given data

The given data is listed below as-

The wavelength of light is $\lambda =2.40 \mathrm{pm}$

Significance of the photoelectric equation

The photoelectric equation which is used to find the wavelength is given by-

$\frac{hc}{\lambda }=\varphi +K_m$

Here, the work function depends on material and condition of the surface and not on the wavelength of incident light.

To find the wavelength of light scattered at 30.00 from the incident direction.(a)

The change in wavelength is obtained as below when a photon scatters from an electron initially at rest:

$\Delta \lambda =\frac{h}{mc}\left(1-\cos \varphi \right)$

Here, m is the mass of electron and $\varphi$is the scattering angle.

Substitute for, $\frac{h}{mc}=2.43\times {10}^{-12}$, $\varphi =30°$and m=2.43 pm in the above equation.

The wavelength is given by-

$\begin{array}{c}\Delta \lambda =\frac{h}{mc}\left(1-\cos \varphi \right)\\ =\left(2.43 \mathrm{pm}\right)\left(1-\cos 30°\right)\\ =0.326 \mathrm{pm}\end{array}$

Now, the final wavelength is:

$\begin{array}{c}\lambda \text{'}=\lambda +\Delta \lambda \\ =2.4 \mathrm{pm}+0.326 \mathrm{pm}\\ =2.73 \mathrm{pm}\end{array}$

Thus,the wavelength of light scattered at 30.0⁰ from the incident direction is 2.73 pm.

To find the wavelength of light scattered at 120.00 from the incident direction.(b)

The change in wavelength is obtained as below when a photon scatters from an electron initially at rest:

$\Delta \lambda =\frac{h}{mc}\left(1-\cos \varphi \right)$

Here, m is the mass of the electron and $\varphi$is the scattering angle

Thus,the wavelength of light scattered at 120.0⁰ from the incident direction is 6.05 pm.