Question

If the atom $$_{100}F{m^{257}}$$  follows the Bohr model and the radius of $$_{100}F{m^{257}}$$  is $$n$$ times the Bohr radius, then find $$n.$$

A. 100
B. 200
C. 4
D. $$\frac{1}{4}$$  
Answer :   $$\frac{1}{4}$$
Solution :
KEY CONCEPT : For an atom following Bohr’s model, the radius is given by
$${r_m} = \frac{{{r_0}{m^2}}}{Z}$$   where $${r_0}$$ = Bohr’s radius and $$m$$ = orbit number.
For $$Fm,m = 5$$   (Fifth orbit in which the outermost electron is present)
$$\therefore {r_m} = \frac{{{r_0}{5^2}}}{{100}} = n{r_0}\left( {{\text{given}}} \right) \Rightarrow n = \frac{1}{4}$$

Releted MCQ Question on
Modern Physics >> Atoms And Nuclei

Releted Question 1

If elements with principal quantum number $$n > 4$$  were not allowed in nature, the number of possible elements would be

A. 60
B. 32
C. 4
D. 64
Releted Question 2

Consider the spectral line resulting from the transition $$n = 2 \to n = 1$$    in the atoms and ions given below. The shortest wavelength is produced by

A. Hydrogen atom
B. Deuterium atom
C. Singly ionized Helium
D. Doubly ionised Lithium
Releted Question 3

An energy of $$24.6\,eV$$  is required to remove one of the electrons from a neutral helium atom. The energy in $$\left( {eV} \right)$$  required to remove both the electrons from a neutral helium atom is

A. 38.2
B. 49.2
C. 51.8
D. 79.0
Releted Question 4

As per Bohr model, the minimum energy (in $$eV$$ ) required to remove an electron from the ground state of doubly ionized $$Li$$ atom $$\left( {Z = 3} \right)$$  is

A. 1.51
B. 13.6
C. 40.8
D. 122.4

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