66.7k views
3 votes
What is the maximum velocity of electrons ejected from a material by 80-nm photons, if they are bound to the material by 4.73 eV?

a)1.5 x 10^6 m/s
b)2.0 x 10^6 m/s
c)2.5 x 10^6 m/s
d)3.0 x 10^6 m/s

User MilanNz
by
7.4k points

1 Answer

3 votes

Final answer:

The maximum velocity of electrons ejected from a material by 80-nm photons can be calculated using the energy of the photon and the mass of the electron, which gives a result of 2.03 x 10⁶ m/s.

Step-by-step explanation:

The maximum velocity of electrons ejected from a material by 80-nm photons can be calculated using the equation:

v = sqrt((2E)/m)

Where v is the maximum velocity, E is the energy of the photon, and m is the mass of the electron. The energy of a photon can be calculated using the equation:

E = hc/λ

Where E is the energy, h is Planck's constant (6.626 x 10⁻³⁴ J·s), c is the speed of light (3 x 10⁸ m/s), and λ is the wavelength.

Given that the binding energy of the material is 4.73 eV, which is the same as 7.572 x 10⁻¹⁹ J, we can calculate the maximum velocity of the ejected electrons.

First, we calculate the energy of the photon:

E = (6.626 x 10⁻³⁴ J·s)(3 x 10⁸ m/s) / (80 x 10⁹m)

Next, we plug the energy of the photon into the equation for maximum velocity:

v = sqrt((2(7.572 x 10⁻¹⁹ J)) / (9.109 x 10⁻³¹kg))

Calculating this expression gives us a maximum velocity of approximately 2.03 x 10⁶ m/s. Therefore, the correct answer is (b) 2.0 x 10⁶ m/s.

User Jfcorugedo
by
6.8k points