Question

If one takes into account the finite mass of the proton, the correction to the binding energy of hydrogen atom is approximately (mass of proton $=1.60\times {{10}^{-27}}\;kg$, the mass of electron $=9.10\times {{10}^{-31}}kg)$.
$\left( a \right)0.06\%$
$\left( b \right)0.0006\%$
$\left( c \right)0.02\%$
$\left( d \right)0.00\%$

Answer 1

Apply Einstein's mass-energy relation. According to this equation, mass and energy are related to each other. That is, mass and energy can be interconverted. That is mass can be converted into energy and energy can be converted into mass. And the binding energy of Hydrogen atoms is produced due to the mass defect.
Formula used:
According to Einstein’s mass-energy relation,
$E=m{{c}^{2}}$
where E is the energy
m is the mass
c is the velocity of light

Complete step-by-step solution:
The mass and energy are related to each other by Einstein’s mass-energy relation. That is,
$E=m{{c}^{2}}$
where E is the energy
m is the mass
c is the velocity of light
This equation tells that mass and energy can be interconverted. That is mass can be converted into energy and energy can be converted into mass.
Also, we must know that the binding energy of the Hydrogen atom is also produced due to the mass defect.
Consider the equation,
$E=m{{c}^{2}}$
Here the equation says that correction in energy is due to correction in mass. Hence here correction is also equal to the error.
Therefore, error in measurement of energy is due to error in measurement of energy.
$\dfrac{dE}{E}\times 100=\dfrac{dm}{m}\times 100$
Here c is not considered because velocity of light is a constant. Hence, there will be no error in the case of ‘c’.
From this we can find dm, the maximum possible correction is the mass of the electron.
$\dfrac{dE}{E}\times 100=\dfrac{dm}{m}\times 100$
$\Rightarrow \dfrac{dE}{E}\times 100=\dfrac{{{m}_{e}}}{{{m}_{p}}}\times 100$
$\Rightarrow \dfrac{dE}{E}\times 100=\dfrac{9.1\times {{10}^{-31}}}{1.6\times {{10}^{-27}}}$ $\times 100$
$\Rightarrow \dfrac{dE}{E}\times 100=0.000568\times 100$
$\Rightarrow \dfrac{dE}{E}\times 100=0.0568$
$\therefore \dfrac{dE}{E}\times 100\simeq 0.06$

Therefore, option (d) is correct.

Note: Here, according to Einstein’s mass-energy relation the correction in energy is due to correction in mass. Hence here correction is also equal to the error. Therefore, error in the measurement of energy is due to error in the measurement of energy. Here c is not considered because the velocity of light is constant. Hence, there will be no error in the case of ‘c’.