The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

       
1. \(\mathrm I\)   
2. \(\mathrm{II}\)
3. \(\mathrm{III}\)
4. \(\mathrm{IV}\)

What happens whenever a hydrogen atom emits a photon in the Balmer series?

Which source is associated with a line emission spectrum? 

1. Electric fibre 

2. Neon street Sign

3. Red traffic light 

4. Sun

Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr's theory, the spectral lines emitted by hydrogen will be:

1. two 

2. three

3. four 

4. one

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

Energy levels A, B, C of a certain atom correspond to increasing values of energy i.e. $E_A<E_B<E_C.$ If ${\lambda }_1,<mspace\; width="1em"></mspace>{\lambda }_2,<mspace\; width="1em"></mspace>{\lambda }_3$ are the wavelengths of the radiation corresponding to the transitions C to B, B to A and C to A respectively, which of the following relation is correct?

1. ${\lambda }_3={\lambda }_1+{\lambda }_2$ 

2. ${\lambda }_3=\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$

3. ${\lambda }_1+{\lambda }_2+{\lambda }_3=0$ 

4. ${\lambda }_3^2={\lambda }_1^2+{\lambda }_2^2$

The element which has K$\alpha$ X-ray line whose wavelength is 0.18 nm is:

1. Cobalt 

2.  Xenon

3. Copper 

4. Iron

Calculate the highest frequency of the emitted photon in the Paschen series of spectral lines of the Hydrogen atom:

1. $3.7\times {10}^{14}<mspace\; width="1em"></mspace>Hz$ 

2. $9.1\times {10}^{15}<mspace\; width="1em"></mspace>Hz$

3. $10.23\times {10}^{14}<mspace\; width="1em"></mspace>Hz$

4. $29.7\times {10}^{15}<mspace\; width="1em"></mspace>Hz$

Hydrogen atoms are excited from ground state of the principal quantum number 4. Then, the number of spectral lines observed will be

1. 3   

2. 6

3. 5

4. 2