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What is the radius of the 5th orbital in hydrogen?

User Amol Udage
by
6.4k points

2 Answers

7 votes

Answer: 1.32*10^-9

Explanation: I had the same question and here's what I got:

rn = n2(r1)

n = 5

r = 5.29*10^-11

rn = n^2(r1)

= 5^2(5.29*10^-11)

r = 1.32*10^-9

User Mradziwon
by
6.3k points
3 votes

Answer: 1.32 x 10^-9 m

Step-by-step explanation:

The formula for the radius of nth orbital is:


r_(n)=(n^(2))/(Z) * 0.528 \text { Angstorms }

For hydrogen, Z= 1, so we have:


$$\begin{aligned}r_(n) &=5^(2) * 0.528 \text { Angstorm } \\&=13.2 \text { Angstorms } \\&=13.2 * 10^(-10) \mathrm{~m} \\&=1.32 * 10^(-9) \mathrm{~m}\end{aligned}$$

User Tanzelax
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6.5k points