The given diagram indicates the energy levels of a certain atom. When the system moves from the \(2E\) level to \(E\), a photon of wavelength \(\lambda\) is emitted. What is the wavelength of the photon produced during its transition from the \(\frac{4E}{3}\) level to \(E\)?

          
1. \(\frac{\lambda}{3}\)
2. \(\frac{3\lambda}{4}\)
3. \(\frac{4\lambda}{3}\)
4. \(3\lambda\)

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 $\stackrel{0}{A}$. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is [IIT-JEE 2011]

1. 1215 $\stackrel{0}{A}$ 

2. 1640 $\stackrel{0}{A}$

3. 2430 $\stackrel{0}{A}$ 

4. 4687 $\stackrel{0}{A}$

Energy E of a hydrogen atom with principal quantum number n is given by $E=\frac{-13.6}{n^2}eV$.  The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen, is approximately [2004]

1. 1.5 eV 

2. 0.85 eV 

3. 3.4 eV 

4. 1.9 eV

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