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4 years ago

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Consider the hydrogen atom. How does the energy difference between adjacent orbit radii change as the principal quantum number i

ncreases?

1 answer:

Kisachek [45]4 years ago

5 0

Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

$E=-E_0 \frac{1}{n^2}$

where

$E_0 = 13.6 eV$ is a constant

n is the level number

We can write therefore the energy difference between adjacent levels as

$\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})$

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

$\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV$

While the difference between the levels n=2 and n=3 is

$\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV$

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

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A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

A)

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

$A\propto N$

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is $m_1 = 1 g$ , the activity is $A_1=16 count/min$

- When the mass is $m_2=20 g$ , the activity is $A_2$

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

B)

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

$A_0$ is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

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$A_0 = 320 count/min$ (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

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C)

We know that a sample of living wood has an activity of

$A=16 count/min$ per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

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Moreover, here we have a sample of 5 g, with current activity of $A=20 count/min$ : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

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- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

$t=2t_{1/2}=2(5700)=11,400 y$

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