Atoms and Nuclei: JEE Main Physics Question with Solution
Experimentally it is found that 12.8 eV energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is x9×10−10 m. The value of the x is ______.
(1eV=1.6×10−19J,4π∈01=9×109 Nm2/C2 and electronic charge=1.6×10−19J C)
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Hint 1 of 3
What formula connects the total binding energy E required to ionize a hydrogen atom to the electron's orbital radius r?
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Correct answer
The value of x is 16.
Option analysis
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Solution
StepWorking
01given
Binding energy Eb=12.8 eV, 1 eV=1.6×10−19 J, k=4πε01=9×109 N⋅m2/C2, e=1.6×10−19 C, orbital radius r=x9×10−10 m.
02find
The value of the integer x.
03visualise
An electron moves in a circular orbit of radius r around a proton. The electrostatic attraction provides centripetal force, giving total mechanical energy E=−2rke2. The energy required to separate them completely to infinity is the binding energy Eb=−E=2rke2.
04strategise
Equate the binding energy expression in Joules to the given value: Eb=2rke2=12.8×(1.6×10−19) J. Solve for r and match with the form x9×10−10.
05execute
Substituting the values:
2r9×109×(1.6×10−19)2=12.8×1.6×10−19r=2×12.89×109×1.6×10−19=25.6/1.69×10−10=169×10−10
Thus, x=16.
✓verify
Radius r=169×10−10 m=0.5625A˚, which is extremely close to the Bohr ground state radius (0.53A˚) corresponding to 13.6 eV, confirming the result is physically sound.
Hints that build this answer step by step
What formula connects the total binding energy E required to ionize a hydrogen atom to the electron's orbital radius r?
E=8πε0re2
Using E=12.8 eV=12.8×1.6×10−19 J, what is the radius r in terms of the given parameters?
r=2×(12.8×1.6×10−19)(9×109)(1.6×10−19)2=169×10−10 m
Comparing r=169×10−10 m with the given form x9×10−10 m, what is the value of x?
Why is the binding energy ke2/(2r) instead of ke2/r?
The binding energy is the work required to remove the electron to infinity, which equals −Etotal. Since total energy E=K+U=21(rke2)−rke2=−2rke2, the separation energy is +2rke2.