
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. $1.2nm$
B. $1.2 \times {10^{ - 3}}nm$
C. $1.2 \times {10^{ - 6}}nm$
D. $1.2 \times {10^1}nm$
Answer
154.2k+ views
Hint: Equate the energy given to remove the proton from its nucleus with the energy of the photon. Energy of a photon is inversely proportional to its wavelength.
Complete answer:
Energy required to remove proton from its nucleus $ = 1MeV = {10^6}eV$
If $\lambda $ is the wavelength of the photon needed to remove the proton, then its energy (in Joules) is given by following relation,
$ \Rightarrow E = \dfrac{{hc}}{\lambda }$, where $h = 6.63 \times {10^{ - 34}}Js$ is Planck’s constant and $c = 3 \times {10^8}m/s$ is the speed of electromagnetic radiation in vacuum.
But, $1eV = 1.66 \times {10^{ - 19}}J$.
To remove the nucleus from its proton, the photon should carry energy equal to one required as mentioned before. Therefore, substituting the values of the constants, we get,
$
\Rightarrow E = \dfrac{{hc}}{\lambda } = {10^6} \times (1.66 \times {10^{ - 19}}) \\
\Rightarrow \lambda = \dfrac{{hc}}{{1.66 \times {{10}^{ - 13}}}} = \dfrac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.66 \times {{10}^{ - 13}}}} \\
\Rightarrow \lambda = 1.2 \times {10^{ - 12}}m \\
$
Now, if we check the options then we find that the units are in $nm = {10^{ - 9}}m$.
So, $\lambda = 1.2 \times {10^{ - 9}} \times {10^{ - 3}}m = 1.2 \times {10^{ - 3}}nm$.
Comparing the options with the final result, we can conclude that option B is correct.
Note: Generally, the value of constants are provided in most of the examinations you will come across in future, but it is still advisable to remember some constants, like the ones used above along with the units. Be careful with powers of 10 while multiplication and division.
Complete answer:
Energy required to remove proton from its nucleus $ = 1MeV = {10^6}eV$
If $\lambda $ is the wavelength of the photon needed to remove the proton, then its energy (in Joules) is given by following relation,
$ \Rightarrow E = \dfrac{{hc}}{\lambda }$, where $h = 6.63 \times {10^{ - 34}}Js$ is Planck’s constant and $c = 3 \times {10^8}m/s$ is the speed of electromagnetic radiation in vacuum.
But, $1eV = 1.66 \times {10^{ - 19}}J$.
To remove the nucleus from its proton, the photon should carry energy equal to one required as mentioned before. Therefore, substituting the values of the constants, we get,
$
\Rightarrow E = \dfrac{{hc}}{\lambda } = {10^6} \times (1.66 \times {10^{ - 19}}) \\
\Rightarrow \lambda = \dfrac{{hc}}{{1.66 \times {{10}^{ - 13}}}} = \dfrac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.66 \times {{10}^{ - 13}}}} \\
\Rightarrow \lambda = 1.2 \times {10^{ - 12}}m \\
$
Now, if we check the options then we find that the units are in $nm = {10^{ - 9}}m$.
So, $\lambda = 1.2 \times {10^{ - 9}} \times {10^{ - 3}}m = 1.2 \times {10^{ - 3}}nm$.
Comparing the options with the final result, we can conclude that option B is correct.
Note: Generally, the value of constants are provided in most of the examinations you will come across in future, but it is still advisable to remember some constants, like the ones used above along with the units. Be careful with powers of 10 while multiplication and division.
Recently Updated Pages
JEE Main 2022 (June 29th Shift 2) Maths Question Paper with Answer Key

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 29th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 26th Shift 2) Chemistry Question Paper with Answer Key

JEE Main 2022 (June 26th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (June 29th Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Displacement-Time Graph and Velocity-Time Graph for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Electrical Field of Charged Spherical Shell - JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
