Light of wavelength 488 nm is produced by an Argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photo electrons is 0.38 V. Find the work function of the material form which the emitter is made.

Question

Light of wavelength 488 nm is produced by an Argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photo electrons is 0.38 V. Find the work function of the material form which the emitter is made. (A) 1.38 eV (B) 2.55 eV (C) 2.17eV (D) 2.93 eV

Tardigrade Admin · Accepted Answer

E=(h c/λ )=(1242 eV/488)=2.55 eV Km a x=eV0=e× 0.38=0.38 eV Km a x=(h c/λ )-φ0=E-φ0 φ0=E-Km a x=(2 .55 - 0 .38)=2.17 eV

Q. Light of wavelength 488 nm is produced by an Argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photo electrons is 0.38 V. Find the work function of the material form which the emitter is made.

$1.38 e V$

$2.55 e V$

$2.17 e V$

$2.93 e V$

$E = \frac{h c}{λ} = \frac{1242 e V}{488} = 2.55 e V$
$K_{ma x} = e V_{0} = e \times 0.38 = 0.38 e V$
$K_{ma x} = \frac{h c}{λ} - ϕ_{0} = E - ϕ_{0}$
$ϕ_{0} = E - K_{ma x} = (2.55 - 0.38) = 2.17 e V$

Q. Light of wavelength 488 nm is produced by an Argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photo electrons is 0.38 V. Find the work function of the material form which the emitter is made.

1.38eV

2.55eV

2.17eV

2.93eV

E=λhc​=4881242eV​=2.55eV Kmax​=eV0​=e×0.38=0.38eV Kmax​=λhc​−ϕ0​=E−ϕ0​ ϕ0​=E−Kmax​=(2.55−0.38)=2.17eV

$1.38 e V$

$2.55 e V$

$2.17 e V$

$2.93 e V$

$E = \frac{h c}{λ} = \frac{1242 e V}{488} = 2.55 e V$
$K_{ma x} = e V_{0} = e \times 0.38 = 0.38 e V$
$K_{ma x} = \frac{h c}{λ} - ϕ_{0} = E - ϕ_{0}$
$ϕ_{0} = E - K_{ma x} = (2.55 - 0.38) = 2.17 e V$