The energy of ground electronic state of hydrogen atom is -13.6 eV.  The energy of the first excited state will be [1997]

1. -54.4 eV 

2. -27.2 eV

3. -6.8 eV 

4. -3.4 eV

The total energy of an electron in the first excited state of hydrogen is about -3.4 eV. Its kinetic energy in this state is [2005]

1. -3.4 eV

2. -6.8 eV 

3. 6.8 eV 

4. 3.4 eV

When hydrogen atom is in its first excited level, its radius is [1997]

1. four times, its ground state radius

2. twice, its ground state radius

3. same as its ground state radius

4. half of its ground state radius

The ionisation potential of helium atom is 24.6 volt, the energy required to ionise it will be

1. 24.6 eV 

2. 24.6 volt 

3. 13.6 volt 

4.13..6 eV

The ionization energy of 10 times ionized sodium atom is:

1. 13.6 eV 

2. $13.6\times 11 eV$ 

3. $\frac{13.6}{11}eV$ 

4. $13.6\times {\left(11\right)}^2 eV$

The ionisation energy of hydrogen atom is 13.6 eV, the ionisation energy of helium atom would be [1988]

1. 13.6 eV

2. 27.2 eV 

3. 6.8 eV

4. 54.4 eV

The photon radiated from hydrogen corresponding to the second line of Lyman series is absorbed by a hydrogen-like atom X in the second excited state.   As a result the hydrogen-like atom X makes a transition to $n^{th}$ orbit. Then:

1. X = $H{e}^{+}, n=4$ 

2. X = $L{i}^{++}$, n = 6

3. X = $H{e}^{+}, n=9$ 

4. X = $L{i}^{++}$, n = 9

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

1. $v=\frac{36 C}{5 R}$ 

2. v = $\frac{CR}{6}$

3. v = $\frac{5 RC}{36}$

4. v  = $\frac{6 C}{R}$

In an experiment to determine the e/m value for an electron using Thomson's method the electrostatic deflection plates were 0.01 m apart and had a potential difference of 200 volts applied. Then the electric field strength between the plates is 

1. $1\times {10}^4 V/m$ 

2. $2\times {10}^4 V/m$

3. $4\times {10}^4 V/m$ 

4. $5\times {10}^5 V/m$