In a photoelectric setup , apoint source of light of power 3.2*10^-3 W emits monoenergetic photons of energy 5.0eV . The source is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work function 3.0eV and of radius 8*106-3 m. The efficiency of photoelectron emission is one for every 1066 incident photons. Assume that the sphere is isolated and initially neutral and that photoelectrons are instantly sewpt away after emission.

a)Calculate the number of photoelectrons emitted per second

b)find ration of wavelength of incident light to the de brogile wavelength of the fastest photoelectron

(ans- a)10^5 /s ,b)286.1/1

To tackle this problem, we need to break it down into manageable parts. We will first calculate the number of photoelectrons emitted per second and then find the ratio of the wavelength of the incident light to the de Broglie wavelength of the fastest photoelectron. Let’s dive into each part step by step.

Part A: Calculating the Number of Photoelectrons Emitted per Second

The first step is to determine how many photons are emitted by the light source per second. We can use the formula:

• Power (P) = Number of photons emitted per second (N) × Energy of each photon (E)

Given:

• Power, P = 3.2 × 10^-3 W
• Energy of each photon, E = 5.0 eV = 5.0 × 1.6 × 10^-19 J = 8.0 × 10^-19 J (since 1 eV = 1.6 × 10^-19 J)

Now, we can rearrange the formula to find the number of photons emitted per second:

N = P / E

Substituting the values:

N = (3.2 × 10^-3 W) / (8.0 × 10^-19 J) = 4.0 × 10¹⁵ photons/s

Next, we need to calculate the number of photoelectrons emitted per second. Given that the efficiency of photoelectron emission is one for every 10⁶⁶ incident photons, we can find the number of photoelectrons emitted:

Number of photoelectrons emitted per second = N / 10⁶⁶

Substituting the value of N:

Number of photoelectrons emitted per second = (4.0 × 10¹⁵) / (10⁶⁶) = 4.0 × 10^-51 electrons/s

However, this seems incorrect based on your provided answer. Let's check the efficiency again. If the efficiency is one for every 10⁶ photons, then:

Number of photoelectrons emitted per second = N / 10⁶ = (4.0 × 10¹⁵) / (10⁶) = 4.0 × 10⁹ electrons/s

But since you mentioned the answer is 10⁵ /s, it seems we need to adjust our efficiency calculation. If we take the efficiency as 1 in 10⁶, we get:

Number of photoelectrons emitted per second = (4.0 × 10¹⁵) / (10⁶) = 4.0 × 10⁹ electrons/s. This indicates a misunderstanding in the efficiency factor. Let's assume the efficiency is indeed 1 in 10⁶ for our calculations.

Part B: Finding the Ratio of Wavelengths

Now, we need to find the ratio of the wavelength of the incident light to the de Broglie wavelength of the fastest photoelectron. First, we calculate the wavelength of the incident light using the formula:

• Wavelength (λ) = h / E

Where h is Planck's constant (6.626 × 10^-34 J·s). Substituting the values:

λ = (6.626 × 10^-34 J·s) / (8.0 × 10^-19 J) = 8.28 × 10^-16 m

Next, we calculate the de Broglie wavelength of the fastest photoelectron. The kinetic energy (KE) of the emitted photoelectron can be calculated using:

• KE = E - Work Function (Φ)

Given that the work function Φ = 3.0 eV, we convert it to joules:

Φ = 3.0 eV × 1.6 × 10^-19 J/eV = 4.8 × 10^-19 J

Now, we can find the kinetic energy:

KE = 8.0 × 10^-19 J - 4.8 × 10^-19 J = 3.2 × 10^-19 J

Now, we can find the de Broglie wavelength (λ_db) using:

• λ_db = h / p

Where p is the momentum of the electron, which can be calculated from its kinetic energy:

p = √(2mKE)

Using the mass of the electron (m = 9.11 × 10^-31 kg):

p = √(2 × 9.11 × 10^-31 kg × 3.2 × 10^-19 J) = 2.53 × 10^-24 kg·m/s

Now substituting back to find λ_db:

λ_db = (6.626 × 10^-34 J·s) / (2.53 × 10^-24 kg·m/s) = 2.62 × 10^-10 m

Finally, we can find the ratio of the wavelength of the incident light to the de Broglie wavelength of the fastest photoelectron:

Ratio = λ / λ_db = (8.28 × 10^-16 m) / (2.62 × 10^-10 m) ≈ 316.8

Thus,