Question

The most probable radius (in pm) for finding the electron in $H{e^ + }$ is:
A.0.0
B.52.9
C.26.54
D.105.8

Answer 1

Hint: Bohr’s radius is a physical constant which expresses the most probable distance between the electron and the nucleus in a hydrogen atom in ground state. Moreover, the value of this radius is $5.2917721067 \times {10^{ - 11}}m$ .
Formula used:
Bohr’s radius $ = \dfrac{{0.529 \times {{10}^{ - 10}}{n^2}}}{Z}m$
 $ = \dfrac{{52.9{n^2}}}{Z}pm$ Where, n is the number of shell and Z is the atomic number

Complete step by step answer:
Bohr’s model of hydrogen atom was the first atomic model to successfully explain the radiation spectra of atomic hydrogen. Further, the value of the atomic radius was given by the equation:
 $r(n) = {n^2} \times r(1)$ Where, n is the positive integer and r (1) is the smallest allowed radius for the hydrogen atom which is also known as Bohr’s radius.
Now, let’s calculate the most probable radius for finding the electron in $H{e^ + }$ . According to the formula the radius is given by:
r $ = \dfrac{{0.529 \times {{10}^{ - 10}}{n^2}}}{Z}m$
So, atomic number of helium is $2$
And $n = 1$
Now, substitute the values in the given formula
 $ = \dfrac{{52.9{n^2}}}{z}$
 $ = \dfrac{{52.9 \times {1^2}}}{2}$
 =26.45pm
Moreover, according to the Bohr’s model, the energy obtained is always a negative number and the ground state $n = 1$ has the most negative value. This is because the energy of an electron in an orbit is relative to the energy of an electron that is entirely separated from its nucleus, $n = \infty $ and is recognized to have an energy of 0eV.

Hence, option C is correct.

Note:
Although the Bohr’s model is not used in physics but it is highly used in calculating other fundamental physical constants. For example Atomic units, fine structure constant etc. Moreover, this model does not work well for complex atoms. It further could not explain why some spectral lines are more intense than others.