Question

The wavelength of the γ-ray emitted in

is _______ Å. (rounded off to three decimal places)
[Given: h=6.626×10 ⁻³⁴ J s; c=2.998×10⁸ m s⁻¹ ; 1 MeV=1.602×10⁻¹³ J]

Answer 1

Correct Answer: 0.017 - 0.021

The wavelength \( \lambda \) of a photon can be calculated using the energy-wavelength relationship:

E = \( \frac{hc}{\lambda} \)

Where:

• E is the energy of the photon
• h is Planck’s constant \((6.626 \times 10^{-34} \, \text{J.s})\)
• c is the speed of light \((2.998 \times 10^8 \, \text{m/s})\)
• \( \lambda \) is the wavelength

We can rearrange the formula to solve for the wavelength \( \lambda \):

\( \lambda = \frac{hc}{E} \)

Given that the energy of the γ-ray is 0.66 MeV, we first need to convert this to joules:

\(E = 0.66 \, \text{MeV} \times 1.602 \times 10^{-13} \, \text{J/MeV} = 1.0573 \times 10^{-13} \, \text{J}\)

Now, substitute the values into the equation for \( \lambda \):

\(\lambda = \frac{(6.626 \times 10^{-34} \, \text{J.s})(2.998 \times 10^8 \, \text{m/s})}{1.0573 \times 10^{-13} \, \text{J}}\)

\(\lambda = 0.021 \, \text{Å}\)

Thus, the wavelength of the γ-ray is 0.021 Å (rounded to three decimal places).