Question

Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. What could you do to increase the maximum kinetic energy of electrons to 1.5 eV?

Answer 1

To increase the kinetic energy of ejected electrons to 1.5 eV, light with a shorter wavelength than 495 nm needs to be used, ensuring it's above the cutoff wavelength for the material in question, based on the principles of the photoelectric effect.

Step-by-step explanation:

To increase the maximum kinetic energy of electrons ejected by light from 0.5 eV to 1.5 eV, you would need to use light with a higher energy, which corresponds to a shorter wavelength. According to the photoelectric effect, the maximum kinetic energy (KEmax) of the ejected electrons can be determined by the equation KEmax = hf - ϕ, where h is Planck's constant, f is the frequency of the incident light, and ϕ is the work function of the material (the minimum energy required to remove an electron from the surface).

Since the energy of a photon is inversely proportional to its wavelength (λ), as given by the formula E = hc/λ (where c is the speed of light), using light with a shorter wavelength than 495 nm will result in higher energy photons, thus increasing the kinetic energy of the ejected electrons provided that the new wavelength is still above the cutoff wavelength for the particular material.

Answer 2

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Step-by-step explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

`K_(max)=(hc)/(\lambda)-W`

Here h is the Planck's constant, c is the speed of light,
`\lambda` is the wavelength of the light and W the work function of the element:

`W=(hc)/(\lambda)-K_(max)\\W=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(495*10^(-9)m)-0.5eV\\W=2.01eV`

Now, we calculate the wavelength for the new maximum kinetic energy:

`W+K_(max)=(hc)/(\lambda)\\\lambda=(hc)/(W+K_(max))\\\lambda=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(2.01eV+1.5eV)\\\lambda=3.54*10^(-7)m=354*10^(-9)m=354nm`

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.