IIT JEE - NEET

a) -313.6, –34.84

b) -313.6, -78.4

c) -78.4, -34.84

d) -78.4, -19.6

Logic:

Energy of a quantum level or orbit in hydrogen atom in kCal mol^{-1} = -313.6/n^{2}

H_{α} line in Lyman series of hydrogen atomic spectrum corresponds to n_{1}= 1 to n_{2} = 2. It is the first line in Lyman series.

Energy of n = 1 level = -313.6/1^{2} = -313.6 kCal mol^{-1}

Energy of n = 2 level = -313.6/2^{2} = -78.4 kCal mol^{-1}

Conclusion:

The correct option is "b".

(EAMCET 2010-E)

a) n_{2}=∞ to n_{1}=2

b) n_{2}=4 to n_{1}=3

c) n_{2}=2 to n_{1}=1

d) n_{2}=5 to n_{1}=3

**Logic:**

1) Wavelength is inversely proportional to energy. Hence lowest wavelength means, highest energy.

2) Since the energy gap between adjacent levels decreases while move away from the nucleus, the transition involving highest energy is associated with smaller n_{1} value.

**Solution:**

The transition corresponding to n_{2} = 2 to n_{1} = 1 involves smaller n_{1} value and hence is associated with lowest wavelength (highest energy). The spectral line belongs to Lyman series.

Conclusion:

Correct option is "c".

(EAMCET 2005-M)

a) 5hcR/36

b) 4hcR/3

c) 3hcR/4

d) 7hcR/144

**Logic:**

* The wavenumber and energies of spectral lines corresponding to transitions between quantum levels: n_{1} and n_{2} are calculated by using following expressions.

Where:

h = planck's constant

c = velocity of light

R = Rydberg's constant

Z = Atomic number

n_{1}< n_{2}

* For Lyman series: n_{1} = 1 and n_{2} > 1

* Lowest energy corresponds to smaller n_{2} value. i.e. 2. It is the first line in Lyman series.

**Solution:**

Energy of spectral line corresponds to lowest energy in Lyman series for hydrogen atom:

Note: Z = 1 for hydrogen.

Conclusion:

Correct option is "c".

**(new)
Click
here to see 3d Interactive Solved Question paper**

(AIEEE 2003)

a) 3 ----> 2

b) 5 ----> 2

c) 4 ----> 1

d) 2 ----> 5

**Logic and solution:**

* Red end means the spectral line belongs to visible region. Only Balmer series appears in visible region.

* For Balmer series n_{1} = 2.

* Red end represents lowest energy. Hence the third line from this end means n_{2} = 5.

transition |
line number |

2 -----> 3 | 1 |

2 -----> 4 | 2 |

2 -----> 5 | 3 |

Conclusion:

Correct option is "b".

(IIT JEE 2011)

a) n=2 to n=1

b) n=3 to n=2

c) n=4 to n=3

d) n=3 to n=1

**Logic and solution:**

The frequencies and hence the wavenumbers of transitions in H atom and He^{+} ion are equal.

By comparing above terms:

n_{1} = 1 and n_{2} = 2.

Conclusion:

Correct option is "b".

1) What is the wavelength of limiting line in Lyman series?

2) Which spectral series in hydrogen atomic spectrum appears in IR region?

**Jump to Heisenberg's
uncertainty principle - IIT JEE - NEET - IT JAM solved problems **