Question

Wavelength of a $ 1{\text{ keV}} $ is $ {\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{ - 9}}m $ . What is the frequency of $ 1{\text{ MeV}} $ photon?
(A) $ {\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{15}}Hz $
(B) $ {\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{18}}Hz $
(C) $ {\text{2}}{\text{.4}} \times {\text{1}}{{\text{0}}^{20}}Hz $
(D) $ {\text{2}}{\text{.4}} \times {\text{1}}{{\text{0}}^{23}}Hz $

Answer 1

Hint : To solve this question we need to use the Planck’s energy equation for the energy of a photon. Also, we have to use the relation between the frequency and the wavelength of the radiation emitted by the photon.

Formula Used: The formula which are used to solve this question is given by
 $\Rightarrow E = h\nu $ , here $ E $ is the energy of a photon of frequency $ \nu $ , and $ h $ is Planck's constant.
 $\Rightarrow \lambda \nu = c $ , here $ \nu $ is the frequency of an electromagnetic radiation, $ \lambda $ is its wavelength, and $ c $ is its speed.

Complete step by step answer
We know that the energy of a photon is given by
 $\Rightarrow E = h\nu $ ………………………(i)
Also, the frequency of the photon is related to its wavelength by
 $\Rightarrow \lambda \nu = c $
 $ \Rightarrow \nu = \dfrac{c}{\lambda } $ ………………………(ii)
Putting (ii) in (i) we get
 $\Rightarrow E = \dfrac{{hc}}{\lambda } $ ………………………(iii)
According to the question, we have $ {E_1} = 1{\text{ keV = 1}}{{\text{0}}^3}{\text{ eV}} $ , and $ {\lambda _1} = {\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{ - 9}}m $ . Putting these in (iii) we get
 $\Rightarrow {\text{1}}{{\text{0}}^3} = \dfrac{{hc}}{{{\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{ - 9}}}} $
 $ \Rightarrow hc = {\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{ - 6}}{\text{ eV}}m $ ………………………(iv)
Now we have $ {E_2} = 1{\text{ MeV = 1}}{{\text{0}}^6}{\text{ eV}} $ . Let $ {\lambda _2} $ be the corresponding wavelength. Putting these in (iii) we get
 $\Rightarrow {\text{1}}{{\text{0}}^6}{\text{ eV}} = \dfrac{{hc}}{{{\lambda _2}}} $
Substituting (iv) above we get
 $\Rightarrow {\text{1}}{{\text{0}}^6}{\text{ eV}} = \dfrac{{{\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{ - 6}}{\text{ eV}}m}}{{{\lambda _2}}} $
 $ \Rightarrow {\lambda _2} = \dfrac{{{\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{ - 6}}{\text{ eV}}m}}{{{\text{1}}{{\text{0}}^6}{\text{ eV}}}} $
On solving we get
 $\Rightarrow {\lambda _2} = {\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{ - 12}}m $
From (ii) we have
 $\Rightarrow \nu = \dfrac{c}{\lambda } $ ………………………(v)
We know that $ c = 3 \times {10^8}m/s $ . Also substituting (v) above we get
 $\Rightarrow \nu = \dfrac{{3 \times {{10}^8}}}{{{\text{1}}{\text{.24}} \times {\text{1}}{{\text{0}}^{ - 12}}}} $
On solving we get
 $\Rightarrow \nu = {\text{2}}{\text{.4}} \times {\text{1}}{{\text{0}}^{20}}Hz $
Thus, the frequency of $ 1{\text{ MeV}} $ photon is equal to $ {\text{2}}{\text{.4}} \times {\text{1}}{{\text{0}}^{20}}Hz $ .
Hence the correct answer is option C.

Note
The wavelength which is given in this question for the energy of $ 1{\text{ keV}} $ was just given to ease out the calculations. We could have solved this question without using this information. But that would have required much calculation because of conversions of the quantities into their corresponding SI units. So, we have used the given information.