Question

A beam of monochromatic light of wavelength λ ejects photoelectrons from a cesium surface (Φ = 1.9 eV). These photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of λ for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light.

Answer 1

(a) The maximum value of λ for which hydrogen atoms may be ionized is 80 n-m

(b) The maximum value of λ for which hydrogen atoms may get excited from the ground state to the first excited state is  102 n-m

(c) The maximum value of λ for which  the excited hydrogen atoms may emit visible light is 328.04 n-m

Explanation:

It is given that the cesium surface’s work function, ϕ   = 1.9 eV

(a) To ionize a hydrogen atom, the energy required in its ground state,

E = 13.6 eV

From the photoelectric equation of Einstein,

$\frac{\mathrm{hc}}{\lambda}$ = E +  ϕ

Here,

c = Speed of light

h = Constant of Planck

λ= Light’s wavelength

Therefore,

(hc ÷ λ )-  ϕ = E

$\frac{hc}{\lambda} - 1.9 = 13.6$

λ = 80 n-m

(b) From the states  $n_1 = 1$ to $n_2 =2$ , the energy absorbed when the electron is excited

$E_1$ is given by $E_1 = 13.6(\frac{1}{n_1^2} - \frac{1}{n_2^2} )$

$E_1$  = 13.66 × 34

For the photoelectric equation of Einstein,  

Therefore, λ

λ = 102 n-m

(c) There will be an emission by the excited atom from the visible light if an electron jumps to the third orbit from the second orbit, that is, to $n_2$ = 3 from  $n_1$ = 2. This is due to the Balmer series that is in the noticeable region.

Energy  ($E_2$)  = 13.66 × 536

For the photoelectric equation of Einstein,

$\frac{\mathrm{hc}}{\lambda}=\frac{13.6 \times 5}{36}+1.9$

λ = 328.04 n-m

The maximum value of λ for which hydrogen atoms may be ionized is       80 n-m, The maximum value of λ for which hydrogen atoms may get excited from the ground state to the first excited state is  102 n-m , The maximum value of λ for which  the excited hydrogen atoms may emit visible light is 328.04 n-m

Answer 2

Answer:

Above answ..er is correct ................