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# Quantum mechanics :( watch

1. any help with these 2 questions would be awesome as im completely stuck

1. When radiation of wavelength l is incident on a metallic surface, the stopping potential is 4.8 volts.
If the same surface is illuminated with radiation of double the wavelength, then the stopping
potential becomes 1.6 volts. What is the threshold wavelength for the surface?

2. The transmission probability (PT) describes the likelihood of an incident particle, of given
energy, passing through a rectangular barrier of width L. For a particular system, PT = 1x103.
What will be the value of PT when we double the width of the barrier?
2. simultaneous equations are your friend. Might want to use WKB approx for the 2nd one.
3. for the first one imagine a graph of voltage on the y axis and wavelength on the x axis, where the line cuts 0V is the threshold.

Second one I'm guessing times the original probability but I'm not sure. I'll check my notes
4. (Original post by mf2004)
for the first one imagine a graph of voltage on the y axis and wavelength on the x axis, where the line cuts 0V is the threshold.

Second one I'm guessing times the original probability but I'm not sure. I'll check my notes
but i dont know the wavelength to calculate the gradient
5. (Original post by luvin_chelsea)
but i dont know the wavelength to calculate the gradient
work out what the wavelength is in terms of the shortest wavelength
6. (Original post by luvin_chelsea)
any help with these 2 questions would be awesome as im completely stuck

1. When radiation of wavelength l is incident on a metallic surface, the stopping potential is 4.8 volts.
If the same surface is illuminated with radiation of double the wavelength, then the stopping
potential becomes 1.6 volts. What is the threshold wavelength for the surface?

2. The transmission probability (PT) describes the likelihood of an incident particle, of given
energy, passing through a rectangular barrier of width L. For a particular system, PT = 1x103.
What will be the value of PT when we double the width of the barrier?

1) Double the wavelength -> half the frequency

eV = hf - E

4.8e = hf - E ------------------- (1) (For wavelength l)

1.6e = 0.5hf - E (For wavelength 2l)
3.2e = hf - 2E ------------(2)

Take (1) - (2): E = 1.6e
for threshold wavelength, the energy provided is just sufficient to overcome potential:

hf = E
hf = 1.6e
hc/x = 1.6e

Find x.

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