Each of the following questions is followed by four answers. Select the correct answer in each case.
- If 13.6 eV energy is required to ionize the hydrogen atom, then the required energy to remove an electron from n = 2 is
|a) 10.2 eV||b)) 0 eV||c) 3.4 eV||d) 6.8 eV|
Solution: Energy of electron in a quantized number n is given by
Where E0 is the energy of the electron in the first orbit. So put the values
2. For an atom of the hydrogen atom the radius of the first orbit is given by
|(a) h/me2||(b) me/4h2||(c) h2/4π2kme2||(d) h2me2|
Explanation: See comprehensive question 2, radii of the orbit of electron (put n = 1).
3.The Balmer series is obtained when all the transitions of electrons terminate on
|(a)1st orbit||(b) 2nd orbit||(c) 3rd orbit||(d) 4th orbit|
Explanation: See comprehensive question 1.
4. In accordance to Bohr’s theory the kinetic energy of electron is equal to
|(a) ½ (Ze2)/r||(b) Ze2/r||(c) Ze2/r2||(d) ½ (Ze2/r2)|
Explanation: See question 1.
5. According to Bohr’s theory radius of the quantized orbit is given by
|(a) 4π2m/n2h2Ze2||(b)n2h2/4π2mZe2||(c) 4π2mZe2k/n2h2||(d) n2h2Ze2/4π2m|
Hint: See comprehensive question 2.
6. In Bohr’s model of Hydrogen atom, the lowest orbit corresponds to
|(a) Infinite energy||(b) Maximum energy|
|(c) Minimum energy||(d) Zero energy|
7. When an electron goes from a lower to higher orbit its
|K.E increases, P.E decreases||K.E increases, P.E increases|
|K.E decreases, P.E increases||K.E decreases, P.E decreases|
Explanation: The equations for K.E and P.E possessed by the electron in a certain orbit are
(see comprehensive question 2).
When the electron goes to a higher orbit, its radius rn increases. Therefore, the K.E decreases and P.E increases.The negative sign in P.E shows the nature of energy, i-e, the electron is bound to the nucleus.
8. The frequency of X-rays depends upon
|(a) Number of electrons striking the target||(b) Accelerating potential|
|(c) Nature of the target||(d) Both b and c|
9. Target material use in X-rays tube must have following properties.
|(a) High atomic number and high melting point||(b) High atomic number and low melting point|
|(c) Low atomic number and low melting point||(d) High atomic number only|
Hint: High atomic number means heavy nucleus. High melting point means the material is not easily melted with temperature.
10. Laser is a device which can produce
|(a) Intense beam of light||(b) Coherent beam of light|
|(c) Monochromatic beam of light||(d) All of the above|
Explanation: See conceptual question 10.