Ncert Exercises & Solutions

Question 11.24:
In an accelerator experiment on high-energy collisions Of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair Of total energy 10.2 Bev into two $\gamma$-rays of equal energy. What is the wavelength associated with each $\gamma$- ray? (1 BeV = 10⁹ eV).

Hint: $E=\frac{hc}{\lambda }$

Step 1: Find the energy of each $\gamma$-ray.

Total energy of two $\gamma$-rays, E= 10.2 BeV = 10.2 x 10⁹ eV=10.2 x 10⁹ x 1.6 x 10^-19 J.
Hence, the energy of each $\gamma$-ray:

$\begin{aligned} E^{'} & = \frac{E}{2} \\ = \frac{10 .2 \times 1 .6 \times 10^{- 10}}{2} = 8 .16 \times 10^{- 10} J \end{aligned}$

Step 2: Find the wavelength.
Energy is related to wavelength as:

$\begin{aligned} E^{'} & = \frac{hc}{λ} \\ ∴ λ & = \frac{hc}{E^{'}} \\ = \frac{6 .626 \times 10^{- 34} \times 3 \times 10^{8}}{8 .16 \times 10^{- 10}} = 2 .436 \times 10^{- 16} m \end{aligned}$

Therefore, the wavelength associated with each $\gamma$-ray is 2.436 x 10^-16 m.

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