
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with $1\;MeV$ energy is nearly:
A. $1.2\;nm$
B. $1.2 \times 10^{-3}\;nm$
C. $1.2 \times 10^{-6}\;nm$
D. $1.2 \times 10^{1}\;nm$
Answer
585.9k+ views
Hint: From what is given to us we understand that the photon is required to possess a wavelength and a corresponding energy equivalent to the binding energy of the nucleus over the photon. Equate the photon energy to the binding energy and solve to get the wavelength of the photon. Be sure to convert the photon energy parameters to electron volts $eV$ in order to maintain universality in the system of units used.
Formula used:
Energy possessed by a photon $E_{photon} = \dfrac{hc}{\lambda}$
Complete answer:
To remove the proton bound to the nucleus, the photon must have at least the same amount of energy as the binding energy of the proton to the nucleus.
Thus $E_{photon} = BE_{nucleus}$
Now, the energy possessed by a photon is given as $E_{photon} = \dfrac{hc}{\lambda}$,
where h is the Planck’s constant = $(6.625 \times 10^{-34}\;J.s)$, c is the velocity of light(photon)= $(3 \times 10^8\;ms^{-1})$ and $\lambda$ is the wavelength of the photon.
We have $BE_{nucleus} = 1\;MeV = 10^6\;eV$
Now we need to convert $hc$ into $eV$ units.
We have $hc = 6.625 \times 10^{-34} \times 3 \times 10^8\; J.m = 19.875 \times 10^{-26}\;J.m $
Now, we know that $1\;eV = 1.6 \times 10^{-19}\;J$
Therefore, $hc$ in $eV$ will be equal to: $\dfrac{19.875 \times 10^{-26}}{1.6 \times 10^{-19}} = 12.422 \times 10^{-7} \;eV.m = 1242.2 \times 10^{-9}\; eV.m = 1242.2\;eV.nm$
So now, we have
$E_{photon} = BE_{nucleus} \Rightarrow \dfrac{hc}{\lambda} = 10^6$
$\Rightarrow \dfrac{1242.2}{\lambda} = 10^6 \Rightarrow \lambda = \dfrac{1242.2}{10^6} = 1242.2 \times 10^{-6}\;nm = 1.24 \times 10^{-3}\;nm$.
So, the correct answer is “Option B”.
Note:
Remember to always maintain a consistency in the units and conversions that define the parameters given to us. This is important in order to avoid unnecessary discrepancies that arise as a result of the inconsistency in the system of units we have used while computing, which most often results in an incorrect final solution.
Formula used:
Energy possessed by a photon $E_{photon} = \dfrac{hc}{\lambda}$
Complete answer:
To remove the proton bound to the nucleus, the photon must have at least the same amount of energy as the binding energy of the proton to the nucleus.
Thus $E_{photon} = BE_{nucleus}$
Now, the energy possessed by a photon is given as $E_{photon} = \dfrac{hc}{\lambda}$,
where h is the Planck’s constant = $(6.625 \times 10^{-34}\;J.s)$, c is the velocity of light(photon)= $(3 \times 10^8\;ms^{-1})$ and $\lambda$ is the wavelength of the photon.
We have $BE_{nucleus} = 1\;MeV = 10^6\;eV$
Now we need to convert $hc$ into $eV$ units.
We have $hc = 6.625 \times 10^{-34} \times 3 \times 10^8\; J.m = 19.875 \times 10^{-26}\;J.m $
Now, we know that $1\;eV = 1.6 \times 10^{-19}\;J$
Therefore, $hc$ in $eV$ will be equal to: $\dfrac{19.875 \times 10^{-26}}{1.6 \times 10^{-19}} = 12.422 \times 10^{-7} \;eV.m = 1242.2 \times 10^{-9}\; eV.m = 1242.2\;eV.nm$
So now, we have
$E_{photon} = BE_{nucleus} \Rightarrow \dfrac{hc}{\lambda} = 10^6$
$\Rightarrow \dfrac{1242.2}{\lambda} = 10^6 \Rightarrow \lambda = \dfrac{1242.2}{10^6} = 1242.2 \times 10^{-6}\;nm = 1.24 \times 10^{-3}\;nm$.
So, the correct answer is “Option B”.
Note:
Remember to always maintain a consistency in the units and conversions that define the parameters given to us. This is important in order to avoid unnecessary discrepancies that arise as a result of the inconsistency in the system of units we have used while computing, which most often results in an incorrect final solution.
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