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3 parts
a

4.8×10−19 J\text{}4.8 \times 10^{-19} \text{ J}

b

5.64×10−17 J\text{}5.64 \times 10^{-17} \text{ J}

c

7.23×1014 Hz\text{}7.23 \times 10^{14} \text{ Hz}

Step-by-step solution

7 steps

Here are the solutions to your questions:

1. Calculate the quantum of energy for a radiation of frequency 2.5×10142.5 \times 10^{14} Hz.

Step 1: State the formula for the energy of a photon.
The energy of a quantum of radiation (photon) is given by Planck's equation:

E=hνE = h\nu

where EE is the energy, hh is Planck's constant, and ν\nu is the frequency.

Step 2: Substitute the given values into the formula.
Given:
ν=2.5×1014\nu = 2.5 \times 10^{14} Hz
h=6.63×10−34h = 6.63 \times 10^{-34} J s

E=(6.63×10−34 J s)×(2.5×1014 Hz)E = (6.63 \times 10^{-34} \text{ J s}) \times (2.5 \times 10^{14} \text{ Hz})

Step 3: Calculate the energy.

E=16.575×10−20 JE = 16.575 \times 10^{-20} \text{ J} E=1.6575×10−19 JE = 1.6575 \times 10^{-19} \text{ J}

The quantum of energy is 1.66×10−19 J\boxed{\text{}1.66 \times 10^{-19} \text{ J}} (rounded to three significant figures).

2. The work function of a metal is 3.0 eV.

Part a· 2 steps

Express this energy in Joules.

  1. Use the given conversion factor.

    Given:
    Work function (Φ\Phi) = 3.0 eV
    Conversion factor: 1 eV = 1.6×10−191.6 \times 10^{-19} J

    Φ=3.0 eV×1.6×10−19 J1 eV\Phi = 3.0 \text{ eV} \times \frac{1.6 \times 10^{-19} \text{ J}}{1 \text{ eV}}

  2. Calculate the work function in Joules.

    Φ=4.8×10−19 J\Phi = 4.8 \times 10^{-19} \text{ J}

    The work function in Joules is 4.8×10−19 J\boxed{\text{}4.8 \times 10^{-19} \text{ J}}

Part b

5.64×10−17 J\text{}5.64 \times 10^{-17} \text{ J}

The metal surface is illuminated by electromagnetic radiation of wavelength 3.5 nanometers. Determine the maximum kinetic energy (KE) of the photoelectron.

  1. Calculate the energy of the incident photon.

    The energy of an incident photon is given by E=hcλE = \frac{hc}{\lambda}.
    Given:
    Wavelength (λ\lambda) = 3.5 nanometers = 3.5×10−93.5 \times 10^{-9} m
    Planck's constant (hh) = 6.64×10−346.64 \times 10^{-34} J s (using the value provided in this part of the question)
    Speed of light (cc) = 3.0×1083.0 \times 10^8 m/s (standard value)

    E=(6.64×10−34 J s)×(3.0×108 m/s)3.5×10−9 mE = \frac{(6.64 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m/s})}{3.5 \times 10^{-9} \text{ m}} E=19.92×10−263.5×10−9 JE = \frac{19.92 \times 10^{-26}}{3.5 \times 10^{-9}} \text{ J} E≈5.6914×10−17 JE \approx 5.6914 \times 10^{-17} \text{ J}

  2. Calculate the maximum kinetic energy using the photoelectric effect equation.

    The maximum kinetic energy (KEmaxKE_{max}) of the photoelectron is given by:

    KEmax=E−ΦKE_{max} = E - \Phi KEmax=(5.6914×10−17 J)−(4.8×10−19 J)KE_{max} = (5.6914 \times 10^{-17} \text{ J}) - (4.8 \times 10^{-19} \text{ J})

    To subtract, ensure the exponents are the same:

    KEmax=(569.14×10−19 J)−(4.8×10−19 J)KE_{max} = (569.14 \times 10^{-19} \text{ J}) - (4.8 \times 10^{-19} \text{ J}) KEmax=564.34×10−19 JKE_{max} = 564.34 \times 10^{-19} \text{ J} KEmax=5.6434×10−17 JKE_{max} = 5.6434 \times 10^{-17} \text{ J}

    The maximum kinetic energy is 5.64×10−17 J\boxed{\text{}5.64 \times 10^{-17} \text{ J}} (rounded to three significant figures).

Part c

7.23×1014 Hz\text{}7.23 \times 10^{14} \text{ Hz}

Determine the threshold frequency.

  1. State the formula for threshold frequency.

    The threshold frequency (ν0\nu_0) is the minimum frequency of incident radiation required to cause photoemission, and it is related to the work function by:

    Φ=hν0\Phi = h\nu_0 ν0=Φh\nu_0 = \frac{\Phi}{h}

  2. Substitute the work function (in Joules) and Planck's constant.

    Given:
    Φ=4.8×10−19\Phi = 4.8 \times 10^{-19} J
    h=6.64×10−34h = 6.64 \times 10^{-34} J s

    ν0=4.8×10−19 J6.64×10−34 J s\nu_0 = \frac{4.8 \times 10^{-19} \text{ J}}{6.64 \times 10^{-34} \text{ J s}}

  3. Calculate the threshold frequency.

    ν0≈0.72289×1015 Hz\nu_0 \approx 0.72289 \times 10^{15} \text{ Hz} ν0≈7.23×1014 Hz\nu_0 \approx 7.23 \times 10^{14} \text{ Hz}

    The threshold frequency is 7.23×1014 Hz\boxed{\text{}7.23 \times 10^{14} \text{ Hz}} (rounded to three significant figures).

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