Home
Class 12
PHYSICS
In a photoelectric experiment, it was fo...

In a photoelectric experiment, it was found that the stopping potential decreases from 1.85 V to 0.82 V as the wavelength of the incident light is varied from 300nm to 400nm. Calculate the value of the Planck constant from these data.

Text Solution

Verified by Experts

The maximum kinetic energy of a photoelectron is
` (K_max = hc/lambda - varphi)`
and the stopping potential is
` V= (K_max/e)= (hc/lambda e )- (varphi/ e).`
` If V_1, V_2, are the stopping potentials at wavelengths
lambda_1 and lambda_2 respectively,
` V_1 = (hc/lambda_1 e)- (varphi/e)`
` and V_2= (hc/ (lambda_2 e)-(varphi/ e)).`
` This gives, V_1- V_2 = (hc/e) ( 1/lambda_1) - (1 / lambda_2)
` or, h= (e((V_1)-(V_2))/ c(1/lambda_1)-(1/lambda_2))`
` = (e(1.85 V - 0.82V)/c(1/(300 xx (10^-9)m))-(1/400 xx (10^-9)m))`
` = (1.03e V/ ((3 xx (10^8)m (s^-1))(1/12 xx (10^7)(m^-1)))`
` = 4.12 xx (10^-15) eVs.`
Promotional Banner

Topper's Solved these Questions

  • PHOTO ELECTRIC EFFECT AND WAVE PARTICLE DUALITY

    HC VERMA|Exercise question for short answer|13 Videos

  • PHOTO ELECTRIC EFFECT AND WAVE PARTICLE DUALITY

    HC VERMA|Exercise objective I|14 Videos

  • PHOTO ELECTRIC EFFECT AND WAVE PARTICLE DUALITY

    HC VERMA|Exercise Exercise|35 Videos

  • PERMANENT MAGNETS

    HC VERMA|Exercise Exercise|25 Videos

  • PHOTOMETRY

    HC VERMA|Exercise All Questions|44 Videos

Similar Questions

Explore conceptually related problems

In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12xx10^(-15)V s. Calculate the value of Planck's constant.

The electric potential decreases unifromly from 120 V to 80 V as one moves on the x-axis from x=-1 cm to x=+1 cm. The electric field at the origin

When 40 kV is applied across an X-ray tube, X-ray is obtained with a maximum frequency of 9.7 xx 10 ^18 Hz. Calculate the value of Planck constant from these data.

In Young's double slit experiment, the distance of the screen from the two slits in 1m. When a light of wavelength 600nm is allowed to fall on the slits width of the fringes obtained on the screen is 2mm. Calculate the width of the fringe if the wavelength of the incident light is 400nm. Calculate band width in each case if the arrangement is immersed in water of refractive index 1.33.

The stopping potential in a photoelectric experiment is linearly related to the inverse of the wavelength (1/lambda) of the light falling on the cathode. The potential difference applied across an X-ray tube is linearly related to the inverse of the cutoff wavelength (1/lambda) of the X-ray emitted. Show that the slopes of the lines in the two cases are equal and find its value.

Light of wavelength 390 nm is directed at a metal electrode. To find the energy of electrons ejected, an opposing potential difference is established between it and another electrode. The current of photoelectrons from one to the other is stopped completely when the potential difference is 1.10 V. Determine (i) the work function of the metal and (ii) the maximum wavelength of light that can electrons from this metal.

The light from the sun is found to have a maximum intensity near the wavelength of 470nm. Assuming that the surface o f the sun emits as a blackbody, calculate the temperature of the surface of the sun.

Ina double slit experiment the angular width of a fringe is found to be 0.2^(@) on a screen placed 1m away.The wavelength of light is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water of be 4/3

In a Young's double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the centre of the central maximum is 0.20 W m^-2 , what will be the intensity at a point 0.5 cm away from this centre along the width of the fringes ?