An electron in the ground state of hydrogen has an angular momentum $L_1$ and an electron in the first excited state of lithium has an angular momentum $L_2$. Then

1. $L_1=L_2$

2. $L_1=4L_2$

3. $L_2=2L_1$

4. $L_1=2L_2$

The voltage applied to an X-ray tube is 18 kV. The maximum mass of photon emitted by the X-ray tube will be: 

1. $2\times {10}^{-13} \mathrm{kg}$

2. $3.2\times {10}^{-36 }\mathrm{kg}$

3. $3.2\times {10}^{-32} \mathrm{kg}$

4. $9.1\times {10}^{31} \mathrm{kg}$

The force acting on the electron in a hydrogen atom depends on the principal quantum number as:

1. $\mathrm{F}\propto {\mathrm{n}}^2$

2. $\mathrm{F} \propto \frac{1}{{\mathrm{n}}^2}$

3. $\mathrm{F} \propto {\mathrm{n}}^4$

4. $\mathrm{F} \propto \frac{1}{{\mathrm{n}}^4}$

The ratio (in SI units) of magnetic dipole moment to that of the angular momentum of an electron of mass m kg and charge e coulomb in Bohr's orbit of hydrogen atom is:

1. $\frac{e}{2m}$

2. $\frac{e}{m}$

3. $\frac{2e}{m}$

4. none of these 

If the first excitation potential of the hydrogen-like atom is V volt, then the ionization energy of this atom will be:

1. V electron volt

2. $\frac{2\mathrm{V}}{4}\mathit{ }$electron volt
3. $\frac{4\mathrm{V}}{3}$ electron volt

4. cannot be calculated by the given information

Hydrogen atom in a sample is excited to n = 5 state and it is found that photons of all possible wavelengths are present in the emission spectra. The minimum number of hydrogen atoms in the sample would be:

1. 5

2. 6

3. 10

4. infinite

The radius of the first permitted Bohr orbit for the electron in a hydrogen atom equals $0.5 \stackrel{o}{A}$ and its ground state energy equals -13.6 eV. If the electron in the hydrogen atom is replaced by muon $\left({\mu }^{-}\right)$ [ charge same as electron and mass 207$m_e$], the first Bohr radius and ground state energy will be-   ($m_e$ represents mass of electron)

1. 0.53 $\times {10}^{-13} m, -3.6 eV$

2. 25.6 $\times {10}^{-13} m, -2.8 eV$

3. 2.56 $\times {10}^{-13} m, -2.8 keV$

4. 2.56$\times {10}^{-13} m, -13.6 eV$

Due to transitions among its first three energy levels, the hydrogen atom emits radiation at three discrete wavelengths ${\mathrm{\lambda }}_1, {\mathrm{\lambda }}_2 \mathrm{and} {\mathrm{\lambda }}_3\left({\mathrm{\lambda }}_1 < {\mathrm{\lambda }}_2 < {\mathrm{\lambda }}_3\right)$. Then,

1.  ${\lambda }_1={\lambda }_2+{\lambda }_3$

2.  ${\lambda }_1+{\lambda }_2={\lambda }_3$

3.  $\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}$

4.  $\frac{1}{{\lambda }_1}=\frac{1}{{\lambda }_2}+\frac{1}{{\lambda }_3}$

The wavelength of the first Balmer line caused by a transition from the n = 3 level to the n = 2 level in hydrogen is ${\mathrm{\lambda }}_1$. The wavelength of the line caused by an electronic transition from n = 5 to n = 3 is :

1.  $\frac{375}{128}{\lambda }_1$

2.  $\frac{125}{64}{\lambda }_1$

3.  $\frac{64}{125}{\lambda }_1$

4.  $\frac{128}{375}{\lambda }_1$

An electron with kinetic energy E collides with a hydrogen atom in the ground state. The collision will be elastic :

1.  For all values of E

2.  For E < 10.2 eV

3.  For 10.2 eV < E < 13.6 eV only

4.  For 0 < E < 3.4 eV only