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Temperature distribution along a metal bar (a) Lagged and (b) Unlagged, and 3. Photoelectric effect for tungsten (given lambda and KEmax).

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2 parts
b(i)

4.47 eV4.47 \text{ eV}

b(ii)

1.08×1015 Hz1.08 \times 10^{15} \text{ Hz}

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Here are the solutions to your questions.

2. Temperature distribution along a metal bar

For both cases, assume the heated end is at x=0x=0 with temperature THT_H, and the end dipped in melting ice is at x=Lx=L with temperature 0∘C0^\circ \text{C}. The temperature TT is plotted on the y-axis and the distance xx along the bar on the x-axis.

Part a

Lagged and the other end dipped in melting ice

The graph will be a straight line connecting the point (0,TH)(0, T_H) to (L,0∘C)(L, 0^\circ \text{C}). This indicates a uniform temperature gradient along the bar because there is no heat loss to the surroundings, so the rate of heat flow is constant.

Part b(i)

4.47 eV4.47 \text{ eV}

The work function of the tungsten in eV

Unlagged and the other end dipped in melting ice

The graph will be a curve connecting the point (0,TH)(0, T_H) to (L,0∘C)(L, 0^\circ \text{C}). Due to heat loss from the surface of the unlagged bar to the surroundings, the rate of heat flow decreases along the bar. This results in a temperature gradient that is steeper near the heated end and less steep near the cold end. The curve will be concave up (or convex down).

3. Photoelectric effect for tungsten

Given:
Wavelength of light λ=248 nm=248×10−9 m\lambda = 248 \text{ nm} = 248 \times 10^{-9} \text{ m}
Maximum kinetic energy KEmax=8.6×10−20 JKE_{max} = 8.6 \times 10^{-20} \text{ J}

Constants:
Planck's constant h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s}
Speed of light c=3.00×108 m/sc = 3.00 \times 10^8 \text{ m/s}
Charge of an electron e=1.602×10−19 Ce = 1.602 \times 10^{-19} \text{ C} (1 eV=1.602×10−19 J1 \text{ eV} = 1.602 \times 10^{-19} \text{ J})

The photoelectric equation is KEmax=Ephoton−ΦKE_{max} = E_{photon} - \Phi, where Ephoton=hcλE_{photon} = \frac{hc}{\lambda} and Φ\Phi is the work function.

  1. Calculate the energy of the incident photon (EphotonE_{photon}).

    Ephoton=hcλE_{photon} = \frac{hc}{\lambda} Ephoton=(6.626×10−34 J s)(3.00×108 m/s)248×10−9 mE_{photon} = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{248 \times 10^{-9} \text{ m}} Ephoton=19.878×10−26248×10−9 JE_{photon} = \frac{19.878 \times 10^{-26}}{248 \times 10^{-9}} \text{ J} Ephoton=8.015×10−19 JE_{photon} = 8.015 \times 10^{-19} \text{ J}

  2. Calculate the work function (Φ\Phi) in Joules.

    Φ=Ephoton−KEmax\Phi = E_{photon} - KE_{max} Φ=(8.015×10−19 J)−(8.6×10−20 J)\Phi = (8.015 \times 10^{-19} \text{ J}) - (8.6 \times 10^{-20} \text{ J}) Φ=(80.15×10−20 J)−(8.6×10−20 J)\Phi = (80.15 \times 10^{-20} \text{ J}) - (8.6 \times 10^{-20} \text{ J}) Φ=71.55×10−20 J\Phi = 71.55 \times 10^{-20} \text{ J} Φ=7.155×10−19 J\Phi = 7.155 \times 10^{-19} \text{ J}

  3. Convert the work function from Joules to electron volts (eV).

    ΦeV=7.155×10−19 J1.602×10−19 J/eV\Phi_{\text{eV}} = \frac{7.155 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} ΦeV=4.466 eV\Phi_{\text{eV}} = 4.466 \text{ eV}

    The work function of tungsten is 4.47 eV\boxed{4.47 \text{ eV}}.

Part b(ii)

1.08×1015 Hz1.08 \times 10^{15} \text{ Hz}

The threshold frequency of the tungsten

The threshold frequency (f0f_0) is related to the work function by Φ=hf0\Phi = hf_0.

  1. Calculate the threshold frequency using the work function in Joules.

    f0=Φhf_0 = \frac{\Phi}{h} f0=7.155×10−19 J6.626×10−34 J sf_0 = \frac{7.155 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} f0=1.080×1015 Hzf_0 = 1.080 \times 10^{15} \text{ Hz}

    The threshold frequency of tungsten is 1.08×1015 Hz\boxed{1.08 \times 10^{15} \text{ Hz}}.

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