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Theory For Notes Making

Practice Questions (Basic Level)

Practice Questions (Mains Level)

Practice Questions (Advance Level)

On Line Test (JEE Main Level)

On Line Test (JEE Advance Level)

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Theory For Making Notes

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Practice Questions (Basic Level)

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Practice Questions (JEE Main Level)

1.

If an electron in an hydrogen atom jumps from an orbit ni = 3 to an orbit with level nf = 2, the frequency of the emitted radiation is

(a) f=\frac{{36c}}{{5R}}                                     

(b) f=\frac{{cR}}{6}                                             

(c) f=\frac{{5Rc}}{{36}}                                     

(d) f=\frac{{6c}}{R}.

Ans (c)

2.

When an electron jumps from an orbit of higher energy E1 to an orbit of lower energy E2, the frequency of radiation occurring is given by

(a) \frac{{({{E}_{1}}-{{E}_{2}})}}{h}              

(b) \frac{{({{E}_{1}}+{{E}_{2}})}}{h}             

(c) \frac{{{{{({{E}_{1}}-{{E}_{2}})}}^{2}}}}{h}    

(d) \frac{{({{E}_{1}}-{{E}_{2}})}}{{{{h}^{2}}}}.

Ans  (a)

3.

The first member of Balmer series of H-atom has a wavelength 6561Å . The wavelength of second member will be

(a) 6860 Å                                       

(b) 5860 Å                                       

(c) 4860 Å                                       

(d) 3860 Å

Ans (c)

4.

Energy levels A, B, C of a certain atom correspond to increasing values of energy i.e., EA< EB< EC. If l1, l2, l3 are the wave lengths of radiation corresponding to the transitions C to B, B to A and C to A respectively, which of the following relations is correct?

(a) {{\lambda }_{3}}={{\lambda }_{1}}+{{\lambda }_{2}}                        

(b) {{\lambda }_{3}}=\frac{{{{\lambda }_{1}}\,{{\lambda }_{2}}}}{{{{\lambda }_{1}}+{{\lambda }_{2}}}     </span></p><p><span style="color: #993300;">(c) {{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}=0                   

(d) \lambda _{3}^{2}=\lambda _{1}^{2}+\lambda _{2}^{2}.

Ans (b)

5.

In which of the following transitions will the wavelength be minimum in the case of hydrogen atom?

(a) n = 5 to n = 4                           

(b) n = 4 to n = 3                          

(c) n = 3 to n = 2                           

(d) n = 2 to n = 1

Ans (d)

6.

The wave number of energy emitted when electron comes from fourth to second orbit in hydrogen is 20397 cm–1. The wave number of energy for same transition in helium is

(a) 5099 cm–1                                 

(b) 20497 cm–1                                              

(c) 40994 cm–1                               

(d) 81588 cm–1

Ans (d)

7.

The wavelength of first line of Balmer series of hydrogen is 6562.8Å. Calculate the ionization potential of hydrogen atom.

(a) 13.26 V

(b) 14.22 V

(c) 12.33 V

(d) 13.62 V

Ans (d)

8.

A hydrogen atom is in an excited state of principle quantum number n. It emits a photon of wavelength \displaystyle \lambda when returns to the ground state. The value of n is (R = Rydberg constant)

(a) \sqrt{{\lambda R(\lambda R-1)}}   

(b) \sqrt{{\frac{{(\lambda R-1)}}{{\lambda R}}}}                     

(c) \sqrt{{\frac{{\lambda R}}{{\lambda R-1}}}}                        

(d) \sqrt{{\lambda (R-1)}}

Ans (c)

9.

An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The minimum kinetic energy of colliding electron is  

(a) 10.2 eV                             

(b) 1.9 eV                               

(c) 12.09 eV                           

(d) 13.6 eV

Ans (c)

10.

In the hydrogen atom spectrum, l3–1 and l2–1 represent wavelengths emitted due to transition from second and first excited states to the ground state respectively. The value of \frac{{{{\lambda }_{{3-1}}}}}{{{{\lambda }_{{2-1}}}}} is

(a) \frac{{27}}{{32}}                      

(b) \frac{{32}}{{27}}

(c) \frac{4}{9}                                  

(d) \frac{9}{4}

Ans (a)

11.

Electrons in hydrogen atom revolve in radius 0.53 Å (in ground state). Due to collision, electron starts revolving in radius of 4.77 Å. Change in angular momentum of the electron will be equal to

(a)  2.11 × 10–36 kg m2/sec                     

(b) 4.22 × 10–30 g m2/sec

(c) 2.11 × 10–27 g cm2/sec                      

(d) 4.22 × 10–36 kg m2/sec

Ans (c)

12.

A gas of hydrogen atoms in their ground state is bombarded by electrons with kinetic energy 12.5 eV. What emitted wavelengths would you expect to see?

(a) 102 nm;  122 nm;  238 nm  

(b) 102 nm;  134 nm;  600 nm  

(c) 102 nm;  122 nm;  653 nm  

(d) 104 nm;  126 nm;  640 nm  

Ans (c)

13.

Calculate the average binding energy per nucleon of the following nuclei:

(a)

  _{{20}}^{{40}}Ca

Mass of  _{{20}}^{{40}}Ca nucleus = 39.9626 u        

(a) 8.22 MeV 

(b) 8.35 MeV 

(c) 8.90 MeV

(d) 8.65 MeV

 Ans (d)

(b)

_{{79}}^{{197}}Au

Mass of _{{79}}^{{197}}Au nucleus = 196.9666 u

(a) 8.00 MeV

(b) 5.37 MeV

(c) 7.65 MeV

(d) 8.03 MeV

Ans (d)

Practice Questions (JEE Advance Level)

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