We read your photo as:

This one is physics

Physics

ScanSolve AI Answer

Step-by-step solution

6 steps

Here's the solution to your physics questions:

4.1.1)
The photoelectric effect indicates the particle nature (or quantum nature) of light, suggesting that light consists of discrete packets of energy called photons.

4.1.2)
The maximum speed of the ejected electrons will REMAIN THE SAME.
Reason: The maximum kinetic energy (and thus maximum speed) of the ejected electrons depends on the frequency (or wavelength) of the incident light and the work function of the metal, not its intensity. Increasing the intensity of light only increases the number of photons, which in turn increases the number of ejected electrons, but not the energy of individual photons or the maximum kinetic energy of each electron.

4.1.3)

  1. Convert the wavelength of incident light from nanometers to meters.

    λ=420 nm=420×10−9 m\lambda = 420 \text{ nm} = 420 \times 10^{-9} \text{ m}

  2. Calculate the energy of the incident photons using the formula E=hcλE = \frac{hc}{\lambda}.

    E=(6.63×10−34 J⋅s)(3.0×108 m⋅s−1)420×10−9 mE = \frac{(6.63 \times 10^{-34} \text{ J}\cdot\text{s})(3.0 \times 10^8 \text{ m}\cdot\text{s}^{-1})}{420 \times 10^{-9} \text{ m}} E=1.989×10−25 J⋅m420×10−9 mE = \frac{1.989 \times 10^{-25} \text{ J}\cdot\text{m}}{420 \times 10^{-9} \text{ m}} E=4.7357×10−19 JE = 4.7357 \times 10^{-19} \text{ J}

  3. Calculate the maximum kinetic energy of the ejected electrons using the formula KEmax=12mevmax2KE_{max} = \frac{1}{2}m_e v_{max}^2.

    KEmax=12(9.11×10−31 kg)(4.76×105 m⋅s−1)2KE_{max} = \frac{1}{2}(9.11 \times 10^{-31} \text{ kg})(4.76 \times 10^5 \text{ m}\cdot\text{s}^{-1})^2 KEmax=12(9.11×10−31 kg)(2.26576×1011 m2⋅s−2)KE_{max} = \frac{1}{2}(9.11 \times 10^{-31} \text{ kg})(2.26576 \times 10^{11} \text{ m}^2\cdot\text{s}^{-2}) KEmax=1.0308×10−19 JKE_{max} = 1.0308 \times 10^{-19} \text{ J}

  4. Use the photoelectric equation E=W0+KEmaxE = W_0 + KE_{max} to find the work function (W0W_0) of the metal.

    W0=E−KEmaxW_0 = E - KE_{max} W0=(4.7357×10−19 J)−(1.0308×10−19 J)W_0 = (4.7357 \times 10^{-19} \text{ J}) - (1.0308 \times 10^{-19} \text{ J}) W0=3.7049×10−19 JW_0 = 3.7049 \times 10^{-19} \text{ J}

  5. Calculate the threshold wavelength (λ0\lambda_0) using the work function W0=hcλ0W_0 = \frac{hc}{\lambda_0}.

    λ0=hcW0\lambda_0 = \frac{hc}{W_0} λ0=(6.63×10−34 J⋅s)(3.0×108 m⋅s−1)3.7049×10−19 J\lambda_0 = \frac{(6.63 \times 10^{-34} \text{ J}\cdot\text{s})(3.0 \times 10^8 \text{ m}\cdot\text{s}^{-1})}{3.7049 \times 10^{-19} \text{ J}} λ0=1.989×10−25 J⋅m3.7049×10−19 J\lambda_0 = \frac{1.989 \times 10^{-25} \text{ J}\cdot\text{m}}{3.7049 \times 10^{-19} \text{ J}} λ0=5.368×10−7 m\lambda_0 = 5.368 \times 10^{-7} \text{ m}

  6. Compare the calculated threshold wavelength with the values in the table.

    The calculated threshold wavelength (5.368×10−7 m5.368 \times 10^{-7} \text{ m}) is approximately equal to the threshold wavelength for Sodium (5.37×10−7 m5.37 \times 10^{-7} \text{ m}).
    The metal used in the experiment is Sodium.

    4.2)
    The lines that correctly show transitions that will result in the atom giving off an EMISSION SPECTRUM are Q and S.
    Reason: An emission spectrum is produced when an electron transitions from a higher energy level to a lower energy level, releasing a photon of specific energy. Both Q (E2 to E1) and S (E3 to E1) represent such transitions.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Physics Questions

QWhat is the physicsQLet's begin with physicsQIs this chemistry or physicsQIt's chemistry not physics🥵QI meant in physics
See all Physics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.