The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away ?
Given: The wave length of radiation is 3300 A 0 and the distance of the laser from the photocell is 1 m .
The energy of incident radiation is given as,
E = h c λ
Where, h is the Planck’s constant, c is the speed of light and λ is the wavelength.
By substituting the values in the above equation, we get
E = 6.6 × 10 − 34 × 3 × 10 8 3300 × 10 − 10 = 6 × 10 − 19 J = 6 × 10 − 19 J × 1 eV 1.6 × 10 − 19 J = 3.16 eV
Since, the energy of the incident radiation is greater than the work function of Na and K only, It is less than the work function of Mo and Ni .
Thus, Mo and Ni will not show photoelectric emission.
The distance of the laser from the photocell does not affect the energy of the radiation. So, If the laser is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase. Thus, result will be same as earlier.
Given: The wave length of radiation is
The energy of incident radiation is given as,
Where,
By substituting the values in the above equation, we get
Since, the energy of the incident radiation is greater than the work function of
Thus,
The distance of the laser from the photocell does not affect the energy of the radiation. So, If the laser is brought near the photocells and placed
