Question

light with a wavelength of 415 nm illuminates a metal cathode. the maximum kinetic energy of the emitted electrons is 0.61 ev .

Answer 1

By photoelectric effect, the remaining energy is 3.82 x 10¯¹⁹ joule.

The photoelectric effect occurs when electrons are emitted from metal when the metal is struck by light of certain frequencies. The electrons produced by the photoelectric effect have maximum kinetic energy proportional to stopping potential energy from the metal which can be determined as:

Kmax = e.Vo

where Kmax is maximum kinetic energy and Vo is stopping potential energy.

From the question above, we know that:

Kmax = 0.61 eV = 9.77 x 10 ¯²⁰ joule

λ = 415 nm = 415 x 10¯⁹ m

Light energy is

E = hc / λ

E = 6.63 × 10¯³⁴ . 3 x 10⁸ / ( 415 x 10¯⁹ )

E = 4.79 x 10¯¹⁹ joule

Hence the remaining energy is

ΔE = E - Kmax

ΔE = 4.79 x 10¯¹⁹ - 9.77 x 10 ¯²⁰

ΔE = 3.82 x 10¯¹⁹ joule

Find more on photoelectric effect at: brainly.com/question/21320305

#SPJ4