# Major help needed with nuclear physics question pls

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#1
Technetium-99m is a common medical tracer injected into patients before they have a scan with a gamma camera. Technetium-99m is a gamma emitter with a half-life of about 6 hours. Each gamma ray photon has energy 2.2 × 10^–14 J.
A patient is given a dose with an initial activity of 500 MBq.

Calculate the initial rate of energy emission from the dose of technetium-99m.

I thought you use the equation A=N*lamda

the decay constant lamda can be worked out by ln2/6*3600

A/lamda = N - number of initial nuclei .

Then each of these will decay to give off one gamma photon of energy 2.2*10^-14... I get 0.34 :/

The markscheme simply multiply A and 2.2*10^-14 .. I don't understand why my method is wrong? or even this equation the markscheme used?? If someone could please elaborate i'd great appreciate it
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3 years ago
#2
(Original post by MrToodles4)
Technetium-99m is a common medical tracer injected into patients before they have a scan with a gamma camera. Technetium-99m is a gamma emitter with a half-life of about 6 hours. Each gamma ray photon has energy 2.2 × 10^–14 J.
A patient is given a dose with an initial activity of 500 MBq.

Calculate the initial rate of energy emission from the dose of technetium-99m.

I thought you use the equation A=N*lamda

the decay constant lamda can be worked out by ln2/6*3600

A/lamda = N - number of initial nuclei .

Then each of these will decay to give off one gamma photon of energy 2.2*10^-14... I get 0.34 :/

The markscheme simply multiply A and 2.2*10^-14 .. I don't understand why my method is wrong? or even this equation the markscheme used?? If someone could please elaborate i'd great appreciate it
This all revolves around an understanding of what activity is. It is the (average) number of decays per unit time (per second, if we are measuring it in becquerel). The activity you are given tells you that there are 500 x 10^6 such decays every second, and if each decay results in a photon of energy 2.2 x 10^-14 J, it should be clear why you multiply them together to find the initial rate of energy emission (that is, the amount of energy emitted each second).

Your method does correctly find the initial number of nuclei. But when you multiply this by the energy emitted in a decay, you are finding the total amount of energy that the dose is able to emit, not the amount emitted each second.
1
#3
(Original post by Pangol)
This all revolves around an understanding of what activity is. It is the (average) number of decays per unit time (per second, if we are measuring it in becquerel). The activity you are given tells you that there are 500 x 10^6 such decays every second, and if each decay results in a photon of energy 2.2 x 10^-14 J, it should be clear why you multiply them together to find the initial rate of energy emission (that is, the amount of energy emitted each second).

Your method does correctly find the initial number of nuclei. But when you multiply this by the energy emitted in a decay, you are finding the total amount of energy that the dose is able to emit, not the amount emitted each second.
You are amazing at explaining, thank you so so much I understand now. Would've loved to have you as my physics teacher
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