StudyGPT Just Leveled Up – Say Hello to Knowee! 🚀
Knowee
Questions
Features
Study Tools

If maximum velocity with which an electron can be emitted from a photo cell is 4 × 108 cm/s, the stopping potential is (mass of electron = 9 × 10–31 kg)

Question

🧐 Not the exact question you are looking for?Go ask a question

Solution

To find the stopping potential, we can use the equation:

eV = 1/2 mv^2

where e is the charge of an electron (1.6 × 10^-19 C), V is the stopping potential, m is the mass of the electron (9 × 10^-31 kg), and v is the maximum velocity of the electron (4 × 10^8 cm/s).

First, let's convert the maximum velocity from cm/s to m/s:

v = 4 × 10^8 cm/s = 4 × 10^6 m/s

Now we can substitute the values into the equation:

(1.6 × 10^-19 C) V = 1/2 (9 × 10^-31 kg) (4 × 10^6 m/s)^2

Simplifying the equation:

V = (1/2) (9 × 10^-31 kg) (4 × 10^6 m/s)^2 / (1.6 × 10^-19 C)

V = (1/2) (9 × 10^-31 kg) (16 × 10^12 m^2/s^2) / (1.6 × 10^-19 C)

V = (1/2) (9 × 16) × (10^-31 kg) (10^12 m^2/s^2) / (1.6 × 10^-19 C)

V = (1/2) (144) × (10^-19 kg m^2/s^2) / (1.6 × 10^-19 C)

V = 72 × (10^-19 kg m^2/s^2) / (1.6 × 10^-19 C)

V = 45 × 10^-19 kg m^2/s^2 / C

V = 45 × 10^-19 V

Therefore, the stopping potential is 45 × 10^-19 volts.

This problem has been solved

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.