
The wavelength of a 1KeV photon is\[1.24\times {{10}^{-9}}m\]. What is the frequency of 1MeV photons?
A. $1.24\times {{10}^{15}}Hz$
B. $2.4\times {{10}^{20}}Hz$
C. $1.24\times {{10}^{18}}Hz$
D. $2.4\times {{10}^{23}}Hz$
Answer
573.6k+ views
Hint: Using the photoelectric equation proposed by Einstein and relationship between frequency and wavelength we can get the required frequency.
As per the given data,
Wavelength of ${{10}^{3}}eV=1.24\times {{10}^{-9}}m$
Complete step-by-step answer:
Photoelectric equation:
$E=h\upsilon $ . . . . .. . (1)
The frequency of the photon is given by:
$\upsilon =\dfrac{c}{\lambda }$
Step by step answer:
According to quantum theory, the energy of photons is given by Planck's constant times the frequency of photons. The value of planck's constant is$1.05\times {{10}^{-34}}$.
Planck's constant was proposed by Max Planck.
$E=h\upsilon $
Where,
$\begin{align}
& h=\ Planck's\ constant \\
& \upsilon =\ frequency \\
\end{align}$
The relation between energy and frequency of photons was postulated by Albert Einstein by using Planck's quantum theory.
As we know that the frequency of a photon is equal to the speed of light (photon) upon the wavelength of the photon.
Mathematically,
$\upsilon =\dfrac{c}{\lambda }$ . . . . . . . . … .(2)
So by putting the value of equation 2 in 1 we get
$E=\dfrac{hc}{\lambda }$
As $hc$ are constants:
${{E}_{1}}{{\lambda }_{_{1}}}={{E}_{2}}{{\lambda }_{_{2}}}$
By potting the values as per provided data:
$\begin{align}
& 1\times {{10}^{3}}eV\times 1.24\times {{10}^{-9}}m={{10}^{6}}eV\times {{\lambda }_{2}} \\
& {{\lambda }_{2}}=1.24\times {{10}^{-12}} \\
\end{align}$
From equation (2),
$\begin{align}
& {{\upsilon }_{2}}=\dfrac{3\times {{10}^{8}}}{1.24\times {{10}^{-12}}} \\
& {{\upsilon }_{2}}=2.419\times {{10}^{20}}Hz \\
\end{align}$
Thus, the frequency of 1MeV photon is$2.419\times {{10}^{20}}Hz$. So the correct option which satisfies the given question is option B.
So, the correct answer is “Option B”.
Note: In question, the wavelength is given and frequency is to be determined. Don’t leave the question in between just by finding the wavelength for 1MeV. 1k means 1,000 (10 3) and 1M refers to 1,000,000 (106), don’t get jumbled in zeros.
As per the given data,
Wavelength of ${{10}^{3}}eV=1.24\times {{10}^{-9}}m$
Complete step-by-step answer:
Photoelectric equation:
$E=h\upsilon $ . . . . .. . (1)
The frequency of the photon is given by:
$\upsilon =\dfrac{c}{\lambda }$
Step by step answer:
According to quantum theory, the energy of photons is given by Planck's constant times the frequency of photons. The value of planck's constant is$1.05\times {{10}^{-34}}$.
Planck's constant was proposed by Max Planck.
$E=h\upsilon $
Where,
$\begin{align}
& h=\ Planck's\ constant \\
& \upsilon =\ frequency \\
\end{align}$
The relation between energy and frequency of photons was postulated by Albert Einstein by using Planck's quantum theory.
As we know that the frequency of a photon is equal to the speed of light (photon) upon the wavelength of the photon.
Mathematically,
$\upsilon =\dfrac{c}{\lambda }$ . . . . . . . . … .(2)
So by putting the value of equation 2 in 1 we get
$E=\dfrac{hc}{\lambda }$
As $hc$ are constants:
${{E}_{1}}{{\lambda }_{_{1}}}={{E}_{2}}{{\lambda }_{_{2}}}$
By potting the values as per provided data:
$\begin{align}
& 1\times {{10}^{3}}eV\times 1.24\times {{10}^{-9}}m={{10}^{6}}eV\times {{\lambda }_{2}} \\
& {{\lambda }_{2}}=1.24\times {{10}^{-12}} \\
\end{align}$
From equation (2),
$\begin{align}
& {{\upsilon }_{2}}=\dfrac{3\times {{10}^{8}}}{1.24\times {{10}^{-12}}} \\
& {{\upsilon }_{2}}=2.419\times {{10}^{20}}Hz \\
\end{align}$
Thus, the frequency of 1MeV photon is$2.419\times {{10}^{20}}Hz$. So the correct option which satisfies the given question is option B.
So, the correct answer is “Option B”.
Note: In question, the wavelength is given and frequency is to be determined. Don’t leave the question in between just by finding the wavelength for 1MeV. 1k means 1,000 (10 3) and 1M refers to 1,000,000 (106), don’t get jumbled in zeros.
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