Ncert Exercises & Solutions

Question 11.3: The photoelectric cut-off voltage in a certain experiment is 1.5 V.
What is the maximum kinetic energy of photoelectrons emitted?

Hint: \(K.E._{max}=eV_{0}\)

Step: Find the maximum kinetic energy of photoelectrons emitted.
Photoelectric cut-off voltage, V₀ = 1.5 V
For emitted photoelectrons, the maximum kinetic energy is:
\(K.E._{max}=eV_{0}\)
\(\Rightarrow~K.E._{max}=eV_{0}=1.6\times10^{-19}\times1.5\)
\(\Rightarrow~K.E._{max}=2.4\times10^{-19}~J.\)

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions