Question

Helium atom emits a photon of wavelength 0.1 angstrom . Recoil energy

Answer 1

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Answer 2

The recoil energy of atom due to the emission of photon is $1.98\times 10^{-14}\ J$.

Explanation:

Given that,

Wavelength of the photon, $\lambda=0.1\ A=0.1\times 10^{-10}\ m$

We need to find the recoil energy of atom due to the emission of photon. It is given by :

$E=\dfrac{hc}{\lambda}$

h is Planck's constant and c is the speed of light

$E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{0.1\times 10^{-10}}$

$E=1.98\times 10^{-14}\ J$

So, the recoil energy of atom due to the emission of photon is $1.98\times 10^{-14}\ J$. Hence, this is the required solution.

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