To solve this problem, we need to use the photoelectric effect equation:

E = hf = eV + Φ

where:
E is the energy of the incident light,
h is Planck's constant (6.63 x 10^-34 Js),
f is the frequency of the light,
e is the charge of an electron (1.6 x 10^-19 C),
V is the stopping potential, and
Φ is the work function of the metal.

First, we need to find the frequency of the light. We know that the speed of light (c) is equal to the product of its wavelength (λ) and frequency (f), so we can rearrange to find:

f = c/λ

Substituting the given values:

f = (3 x 10^8 m/s) / (180 x 10^-9 m)

f = 1.67 x 10^15 Hz

Next, we substitute the values of h, f, e, and V into the photoelectric effect equation to find Φ:

E = hf = eV + Φ

Φ = hf - eV

Substituting the values:

Φ = (6.63 x 10^-34 Js)(1.67 x 10^15 Hz) - (1.6 x 10^-19 C)(0.8 V)

Φ = 1.11 x 10^-18 J - 1.28 x 10^-19 J

Φ = 9.82 x 10^-19 J

However, the work function is usually given in electron volts (eV). To convert from joules to eV, we divide by the charge of an electron:

Φ = (9.82 x 10^-19 J) / (1.6 x 10^-19 C)

Φ = 6.14 eV

So, the work function of the metal surface is approximately 6.14 eV.

Question

To solve this problem, we need to use the photoelectric effect equation:

E = hf = eV + Φ

where:
E is the energy of the incident light,
h is Planck's constant (6.63 x 10^-34 Js),
f is the frequency of the light,
e is the charge of an electron (1.6 x 10^-19 C),
V is the stopping potential, and
Φ is the work function of the metal.

First, we need to find the frequency of the light. We know that the speed of light (c) is equal to the product of its wavelength (λ) and frequency (f), so we can rearrange to find:

f = c/λ

Substituting the given values:

f = (3 x 10^8 m/s) / (180 x 10^-9 m)

f = 1.67 x 10^15 Hz

Next, we substitute the values of h, f, e, and V into the photoelectric effect equation to find Φ:

E = hf = eV + Φ

Φ = hf - eV

Substituting the values:

Φ = (6.63 x 10^-34 Js)(1.67 x 10^15 Hz) - (1.6 x 10^-19 C)(0.8 V)

Φ = 1.11 x 10^-18 J - 1.28 x 10^-19 J

Φ = 9.82 x 10^-19 J

However, the work function is usually given in electron volts (eV). To convert from joules to eV, we divide by the charge of an electron:

Φ = (9.82 x 10^-19 J) / (1.6 x 10^-19 C)

Φ = 6.14 eV

So, the work function of the metal surface is approximately 6.14 eV.

Knowee AI · Accepted Answer

To solve this problem, we need to use the photoelectric effect equation:

E = hf = eV + Φ

where:
E is the energy of the incident light,
h is Planck's constant (6.63 x 10^-34 Js),
f is the frequency of the light,
e is the charge of an electron (1.6 x 10^-19 C),
V is the stopping potential, and
Φ is the work function of the metal.

First, we need to find the frequency of the light. We know that the speed of light (c) is equal to the product of its wavelength (λ) and frequency (f), so we can rearrange to find:

f = c/λ

Substituting the given values:

f = (3 x 10^8 m/s) / (180 x 10^-9 m)

f = 1.67 x 10^15 Hz

Next, we substitute the values of h, f, e, and V into the photoelectric effect equation to find Φ:

E = hf = eV + Φ

Φ = hf - eV

Substituting the values:

Φ = (6.63 x 10^-34 Js)(1.67 x 10^15 Hz) - (1.6 x 10^-19 C)(0.8 V)

Φ = 1.11 x 10^-18 J - 1.28 x 10^-19 J

Φ = 9.82 x 10^-19 J

However, the work function is usually given in electron volts (eV). To convert from joules to eV, we divide by the charge of an electron:

Φ = (9.82 x 10^-19 J) / (1.6 x 10^-19 C)

Φ = 6.14 eV

So, the work function of the metal surface is approximately 6.14 eV.

Light of wavelength 180nm strikes a metal target and photoelectrons are emitted. It is noticed that the photocurrent drops to zero at stopping potential of 0.8 V. The work function of the metal surface is

Question

Solution

Similar Questions

Upgrade your grade with Knowee