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Question: The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus wi...

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1  MeV1\;MeV energy is nearly:
A. 1.2  nm1.2\;nm
B. 1.2×10−3  nm1.2 \times 10^{-3}\;nm
C. 1.2×10−6  nm1.2 \times 10^{-6}\;nm
D. 1.2×101  nm1.2 \times 10^{1}\;nm

Explanation

Solution

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 eVeV in order to maintain universality in the system of units used.

Formula used:
Energy possessed by a photon Ephoton=hcλE_{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 Ephoton=BEnucleusE_{photon} = BE_{nucleus}
Now, the energy possessed by a photon is given as Ephoton=hcλE_{photon} = \dfrac{hc}{\lambda},
where h is the Planck’s constant = (6.625×10−34  J.s)(6.625 \times 10^{-34}\;J.s), c is the velocity of light(photon)= (3×108  ms−1)(3 \times 10^8\;ms^{-1}) and λ\lambda is the wavelength of the photon.
We have BEnucleus=1  MeV=106  eVBE_{nucleus} = 1\;MeV = 10^6\;eV
Now we need to convert hchc into eVeV units.
We have hc=6.625×10−34×3×108  J.m=19.875×10−26  J.mhc = 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=1.6×10−19  J1\;eV = 1.6 \times 10^{-19}\;J
Therefore, hchc in eVeV will be equal to: 19.875×10−261.6×10−19=12.422×10−7  eV.m=1242.2×10−9  eV.m=1242.2  eV.nm\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
Ephoton=BEnucleus⇒hcλ=106E_{photon} = BE_{nucleus} \Rightarrow \dfrac{hc}{\lambda} = 10^6
⇒1242.2λ=106⇒λ=1242.2106=1242.2×10−6  nm=1.24×10−3  nm\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.