The total energy of an electron in the ground state of a hydrogen atom is -13.6 eV. The kinetic energy of an electron in the first excited state is: 

• 3.4 eV

• 6.8 eV

• 13.6 eV

• 13.6 eV

A.

3.4 eV

The total energy of an electron in the orbit is equal to negative of its kinetic energy.
 The energy of hydrogen atom when the electron revolves in nth orbit is 
E = -13.6/n² eV

In the ground state: n =1

E = - -13.6/1² = -13.6 eV

For n = 2, E  = -13.6/2² = -3.4 eV

so, the kinetic energy of an electron in the first excited state (i,e, for n = 2 ) is
K = - E = - (-3.4) - 3.4 eV

B.

three

Ionization energy corresponding to ionization potential  = -13.6 eV.
Photon energy incident = 12.1 eV
So, the energy of electron in excited state
    = -13.6 + 12.1 = -1.5 eV
i.e.,      

i.e., energy of electron in excited state corresponds to third orbit.
The possible spectral lines are when electron jumps from orbit 3rd to 2nd; 3rd to 1st and 2nd to 1st. Thus, 3 spectral lines are emitted.

B.

1:-1

In a Bohr orbit of the hydrogen atom Kinetic energy,

$$\begin{gathered} k = \frac{kz{e}^2}{2r_n} \\ Total energy, E = \frac{-kz{e}^2}{2r_n} \end{gathered}$$

so, Kinetic energy: total energy  = 1:-1

C.

4

For the last Balmer Series

B.

change in the intensity of illumination into a change in photoelectric current

In photoelectric effect when monochromatic radiations of suitable frequency fall on the photo-sensitive plate called cathode, the photoelectrons are emitted which get accelerated towards anode. These electrons flow in the outer circuit resulting in the photoelectric current.
Using the incident radiations of a fixed frequency, it is found that the photoelectric current increases linearly with the intensity of incident light as shown in figure. Hence, a photocell employs photoelectric effect to convert change in the intensity of illumination into a change in photoelectric current.