# Example-1

Example-1

In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?

- 5 â†’ 2
- 4 â†’ 1
- 2 â†’ 5
- 3 â†’ 2

Solution (A)

The electron has minimum energy in the first orbit and its energy increases as
Here

*n*increases.*n*represents the number of orbit, i.e., 1st, 2nd, 3rd, â€¦. The third line from the red end corresponds to yellow region, i.e., 5. In order to obtain less energy, electron tends to come to 1st or 2nd orbit. So jump may be involved either 5 â†’ 1 or 5 â†’ 2. Thus, option (a) is correct here.# Example-2

Example-2

The number of

*d*-electrons retained in Fe^{2+}(atomic number of Fe = 26) ion is- 4
- 5
- 6
- 3

Solution (C)

_{26}Fe = 1

*s*

^{2}, 2

*s*

^{2}, 2

*p*

^{6}, 3

*s*

^{2}, 3

*p*

^{6}, 3

*d*

^{6}, 4

*s*

^{2},

Fe

^{2+}= 1*s*^{2}, 2*s*^{2}, 2*p*^{6}, 3*s*^{2}, 3*p*^{6}, 3*d*^{6}The number of

*d*-electrons retained in Fe^{2+}= 6.Therefore, (c) is the correct option.

# Example-3

Example-3

The difference between the wave number of 1st line of Balmer series and the last line of Paschen series for Li

^{2+}ion is:- 4
*R*

Solution (D)

For 1st line of Balmer series,

For last line of Paschen series,

so,

# Example-4

Example-4

Be

^{3+}and a proton are accelerated by the same potential. Their de Broglie wavelengths have the ratio (assume mass of proton = mass of neutron):- 1 : 2
- 1 : 4
- 1 : 1
- 1 :

Solution (D)

Hence,

# Example-5

Example-5

What is the frequency of revolution of electron present in the 2nd Bohrâ€™s orbit of H atom?

- 1.016 Ã— 10
^{16}s^{â€“1} - 4.065 Ã— 10
^{16}s^{â€“1} - 1.626 Ã— 10
^{15}s^{â€“1} - 8.13 Ã— 10
^{16}s^{â€“1}

Solution (D)

Frequency of revolution