Question

Radiation corresponding to the transition n = 4 to n = 2 in hydrogen atoms falls on a certain metal (work function = 2.5 ev). The maximum kinetic energy of the photoelectrons will be: 1. 0.55 ev 2. 2.55 ev 3. 4.45 ev 4. None of these

Answer 1

Answer : The correct option is, (4) None of these because the kinetic energy of photoelectrons is, 0.05 eV

Solution :

First we have to calculate the energy of photoelectrons.

Formula used :

$E=-13.6\times Z^2(\frac{1}{n_2^2}-\frac{1}{n_1^2})$

where,

E = energy of photoelectrons = ?

Z = atomic number of hydrogen = 1

$n_1$ = final energy level = 4

$n_2$ = initial energy level = 2

Now put all the given values in the above formula, we get

$E=-13.6\times (1)^2(\frac{1}{(4)^2}-\frac{1}{(2)^2})$

$E=2.55eV$

Now we have to calculate the kinetic energy of photoelectrons.

Formula used :

$K.E=E-E_o$

where,

$E_o$ = work function = 2.5 eV

E = energy of photon = 2.55 eV

Now put all the given values in the above formula, we get the kinetic energy of photoelectrons.

$K.E=(2.55eV)-(2.5eV)=0.05eV$

Therefore, the kinetic energy of photo-electrons is, 0.05 eV