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    Hi guys,

    Ok so we got given a really hard sheet on radioactivity and I can't do some of the questions- I'd really like some help!

    1) 'Measuring the total number of beta particles emitted per second is difficult. In one experiment, the count from a small sample of potassium chloride with a known number of potassium-40 nuclei was measured. The sample was spread thinly on a tray and the window of a Geiger tube placed directly above the sample.

    Explain how emission of beta particles downwards into the tray can be taken into account.

    Also, explain how fluctuations in the count rate can be avoided.

    Here's another question I couldn't do:'When a beta minus particle is emitted a nucles of calcium-40 is formed. Since argon is a gas before a rock solidifies there is likely to be no argon present. After the rock has solidified, argon-40 from the decay of potassium-40 is trapped and the amount can be used to measure the time since solidification. A stony meteorite, thought to have been created at the same time as the Earth was analysed. For every 100 atoms of potassium-40 that were still present, there were 120 atoms of argon-40 and 970 nuclei of calcium-40. These were both formed from the decay of the postassium-40 initially present.

    Show that the meteorite solidified 4.7 x10^9 years ago.

    Explain what the emission of a gamma photon of wavelength 2.9x10^-13m shows about the argon-40 nucleus. Use a calculation based on the wavelength in your answer.

    c=3x10^8, h=6.6x10^-34

    Help please :eek::confused: these two questions killed me but the rest were fine S

    MUCH appreciated!

    Aso
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    Omg someone pls help me!
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    My thoughts in italics

    [QUOTE=Aso;41843413]
    1) 'Measuring the total number of beta particles emitted per second is difficult. In one experiment, the count from a small sample of potassium chloride with a known number of potassium-40 nuclei was measured. The sample was spread thinly on a tray and the window of a Geiger tube placed directly above the sample.

    Explain how emission of beta particles downwards into the tray can be taken into account.

    Well, how would you set this up so you could make an estimate of the proportion going downwards?


    Also, explain how fluctuations in the count rate can be avoided

    Well, what might cause fluctuations in the count rate? I must admit I don't quite see what this part is getting at


    Here's another question I couldn't do:'When a beta minus particle is emitted a nucles of calcium-40 is formed. Since argon is a gas before a rock solidifies there is likely to be no argon present. After the rock has solidified, argon-40 from the decay of potassium-40 is trapped and the amount can be used to measure the time since solidification. A stony meteorite, thought to have been created at the same time as the Earth was analysed. For every 100 atoms of potassium-40 that were still present, there were 120 atoms of argon-40 and 970 nuclei of calcium-40. These were both formed from the decay of the postassium-40 initially present.

    You need to know the K-40 half life. This is a decay which can lead to either Ar-40 or Ca-40, with the proportion of the two products determined by the relative rates of the two competing decay processes.

    You can get the relative rates by looking at the relative amounts of Ca-40 and Ar-40 in the meteorite. From this, and the half life of K-40, you can use the conventional radioactive decay equation to work out the time required to reach the K-40/Ar-40 ratio measured.



    Show that the meteorite solidified 4.7 x10^9 years ago.

    Explain what the emission of a gamma photon of wavelength 2.9x10^-13m shows about the argon-40 nucleus. Use a calculation based on the wavelength in your answer.

    c=3x10^8, h=6.6x10^-34



    I am guessing this photon is the 1460 keV photon associated with K-40 decay because I can't be bothered to work it out but, other than telling you the Ar-40 has a high energy nuclear excited state, I really don't see what they are getting at
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    [QUOTE=Cora Lindsay;41846810]My thoughts in italics

    (Original post by Aso)
    1) 'Measuring the total number of beta particles emitted per second is difficult. In one experiment, the count from a small sample of potassium chloride with a known number of potassium-40 nuclei was measured. The sample was spread thinly on a tray and the window of a Geiger tube placed directly above the sample.

    Explain how emission of beta particles downwards into the tray can be taken into account.

    Well, how would you set this up so you could make an estimate of the proportion going downwards?


    Also, explain how fluctuations in the count rate can be avoided

    Well, what might cause fluctuations in the count rate? I must admit I don't quite see what this part is getting at


    Here's another question I couldn't do:'When a beta minus particle is emitted a nucles of calcium-40 is formed. Since argon is a gas before a rock solidifies there is likely to be no argon present. After the rock has solidified, argon-40 from the decay of potassium-40 is trapped and the amount can be used to measure the time since solidification. A stony meteorite, thought to have been created at the same time as the Earth was analysed. For every 100 atoms of potassium-40 that were still present, there were 120 atoms of argon-40 and 970 nuclei of calcium-40. These were both formed from the decay of the postassium-40 initially present.

    You need to know the K-40 half life. This is a decay which can lead to either Ar-40 or Ca-40, with the proportion of the two products determined by the relative rates of the two competing decay processes.

    You can get the relative rates by looking at the relative amounts of Ca-40 and Ar-40 in the meteorite. From this, and the half life of K-40, you can use the conventional radioactive decay equation to work out the time required to reach the K-40/Ar-40 ratio measured.



    Show that the meteorite solidified 4.7 x10^9 years ago.

    Explain what the emission of a gamma photon of wavelength 2.9x10^-13m shows about the argon-40 nucleus. Use a calculation based on the wavelength in your answer.

    c=3x10^8, h=6.6x10^-34



    I am guessing this photon is the 1460 keV photon associated with K-40 decay because I can't be bothered to work it out but, other than telling you the Ar-40 has a high energy nuclear excited state, I really don't see what they are getting at
    Thanks for the help! I'm going straight to my teacher tomorrow!
 
 
 
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