Question

The work function of a metal surface is 4.80 × 10-19 J. The maximum speed of the electrons emitted from the surface is vA = 7.7 × 105 m/s when the wavelength of the light is λA. However, a maximum speed of vB = 5.1 × 105 m/s is observed when the wavelength is λB. Find the wavelengths λA and λB.

Answer 1

The maximum kinetic energy of a photoelectron is the difference between the energy of the incident photon and the work function. The wavelengths λA and λB can be calculated using the given equations.

Step-by-step explanation:

The maximum kinetic energy of a photoelectron at the metal surface is given by the difference between the energy of the incident photon and the work function of the metal. The work function is the binding energy of electrons to the metal surface.

For the first case:

1. Given wavelength λA, use the equation:

hc/λA = Φ + (1/2)mvA²

where h is Planck's constant, c is the speed of light, Φ is the work function, and mA is the mass of the electron.

For the second case:

1. Given wavelength λB, use the equation:

hc/λB = Φ + (1/2)mvB²

You can rearrange the equations to solve for the wavelengths λA and λB:

λA = hc/((Φ + (1/2)mvA²))

λB = hc/((Φ + (1/2)mvB²))

Answer 2

`\lambda_A=2.65177* 10^(-7)\ m`

`\lambda_B=3.32344* 10^(-7)\ m`

Step-by-step explanation:

h = Planck's constant =
`6.63* 10^(-34)\ m^2kg/s`

c = Speed of light =
`3* 10^8\ m/s`

m = Mass of electron =
`9.11* 10^(-31)\ kg`

`W_0` = Work function =
`4.8* 10^(-19)\ J`

`v_A` = Velocity of A particle =
`7.7* 10^5\ m/s`

`v_B` = Velocity of B particle =
`5.1* 10^5\ m/s`

The wavelength is given by

`\lambda=(hc)/((1)/(2)mv^2+W_0)`

`\lambda_A=(6.63* 10^(-34)* 3* 10^8)/((1)/(2)9.11* 10^(-31)(7.7* 10^5)^2+4.8* 10^(-19))\\\Rightarrow \lambda_A=2.65177* 10^(-7)\ m`

The wavelength
`\lambda_A=2.65177* 10^(-7)\ m`

`\lambda_B=(6.63* 10^(-34)* 3* 10^8)/((1)/(2)9.11* 10^(-31)(5.1* 10^5)^2+4.8* 10^(-19))\\\Rightarrow \lambda_B=3.32344* 10^(-7)\ m`

The wavelength
`\lambda_B=3.32344* 10^(-7)\ m`