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sattari [20]
3 years ago
9

What frequency corresponds to an absorption line at 460 nm?

Chemistry
1 answer:
blondinia [14]3 years ago
6 0

Answer:

C. 6.52 x 10^14 Hz

Explanation:

hope this helps

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Phosphorus-32 is radioactive and has a half life of 14.3 days. Calculate the activity of a 3.5mg sample of phosphorus-32. Give y
Andreyy89

Answer:

The activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.

Explanation:

The activity of P-32 can be calculated with the following equation:

A = \lambda N   (1)

Where:

N: is the number of atoms of P-32

λ: is the decay constant

We can find the number of atoms of P-32 as follows:

N = \frac{N_{A}*m}{M}  (2)

Where:

N_{A}: is the Avogadro's number = 6.022x10²³ atoms/mol

m: is the mass of P-32 = 3.5x10⁻³ g

M: is the molar mass of the radionuclide (P-32) = 32 g/mol    

Now, the decay constant is given by:

\lambda = \frac{ln(2)}{t_{1/2}}   (3)

Where:

{t_{1/2}}: is the half-life of P-32 = 14.3 days

Finally, we can find the activity of P-32 by entering equations (2) and (3) into (1):

A = \lambda N = \frac{ln(2)}{t_{1/2}}*\frac{N_{A}*m}{M} = \frac{ln(2)}{14.3 d*\frac{24 h}{1 d}*\frac{3600 s}{1 h}}*\frac{6.022 \cdot 10^{23} mol^{-1}*3.5 \cdot 10^{-3} g}{32 g/mol} = 3.7 \cdot 10^{13} dis/s      

Since a becquerel (Bq) is defined as a disintegration (dis) per second, the activity in Bq is:

A = 3.7 \cdot 10^{13} Bq

And, since a Curie (Ci) is 3.7x10¹⁰ Bq, the activity in Ci is:

A = 3.7 \cdot 10^{13} Bq*\frac{1 Ci}{3.7 \cdot 10^{10} Bq} = 1.0 \cdot 10^{3} Ci

Therefore, the activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.  

               

I hope it helps you!

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