Question

a. What is the decay rate for the $2p$state of hydrogen?

b. During what interval of time will $10\%$of a sample of $2p$hydrogen atoms decay?

Short answer

a) The decay rate for the$2p$state of hydrogen is localid="1650313751027" $6.25\times {10}^8{s}^{-1}$

b) The interval of time in which $10\%$of a sample of localid="1650313761541" $2p$hydrogen atoms will decay islocalid="1650313766475" $0.16s$.

Step-by-step solution

Part(a) Step 1: Given information 

We need to find that what is the decay rate for the $2p$state of hydrogen.

Part(a) Step 2: Simplify

To find decay rate , we will use the formula , $r=\frac{1}{\tau }$, where $r$is decay rate and $\tau$is lifetime of excited state for hydrogen.

So we will substitute the values,$r=\frac{1}{1.6\times {10}^{-9}s}=6.25\times {10}^8{s}^{-1}$

Part(b) Step 1: Given information 

We need to find out in which interval of time $10\%$of a sample of $2p$hydrogen atoms will decay .

Part(b) Step 2: Simplify

We have to find interval of time , here we will number of excited atoms , for that we will use this equation , $N_e=0.90N_0$. Let this be equation $1$.

We also have another formula for $N_e$, that is $N_e=N_o{e}^{-t/\tau }$.Let this be equation $2$.

Comparing the equation $1$and $2$, we get $e^{-t/\tau }=0.90$.

By taking logarithm on both sides , we get , $t=-\tau \ln (0.90)$

Substituting the value of $t$and $\tau$, which we know , we will get,

$$\begin{gathered} t=-(1.6\times {10}^{-9}s)\ln (0.90) \\ t=0.16ns \end{gathered}$$