Question

module ba
problem 27.63
is the use of nonrelativistic formulas justified in the bohr atom?
part a
to check, calculate the electrons velocity, v, in terms of c, for the ground state of hydrogen
express your answer using two significant figures.
v = c
submit request answer
part b
calculate \\(\sqrt{1 - v^2/c^2}\\).
express your answer using five significant figures.

Explanation:

Step1: Recall Bohr's model formula for electron velocity

In the Bohr model of the hydrogen atom, the velocity \( v \) of an electron in the \( n \)-th orbit is given by \( v = \frac{\alpha c}{n} \), where \( \alpha \) (fine - structure constant) is approximately \( 1/137 \), and for the ground state, \( n = 1 \). So, \( v=\frac{c}{137}\approx0.007299c \). Rounding to two significant figures, we get \( v\approx0.0073c \) (or \( 7.3\times 10^{- 3}c \)).

Step2: Calculate \( \sqrt{1 - v^{2}/c^{2}} \)

First, we know that \( v = \frac{c}{137} \), so \( \frac{v^{2}}{c^{2}}=\frac{1}{137^{2}}=\frac{1}{18769}\approx5.328\times 10^{-5} \). Then, \( 1-\frac{v^{2}}{c^{2}}=1 - 5.328\times 10^{-5}=0.99994672 \). Taking the square root, \( \sqrt{0.99994672}\approx0.99997 \) (using a calculator for more precise calculation, \( \sqrt{1-\frac{1}{137^{2}}}=\sqrt{\frac{137^{2}-1}{137^{2}}}=\frac{\sqrt{137^{2}-1}}{137}=\frac{\sqrt{18769 - 1}}{137}=\frac{\sqrt{18768}}{137}\approx\frac{136.9963}{137}\approx0.999973 \) when calculated more accurately. Rounding to five significant figures, we get \( 0.99997 \).

Answer:

Part A: \( v = \boxed{0.0073}c \) (or \( 7.3\times10^{-3}c \))
Part B: \( \sqrt{1 - v^{2}/c^{2}}=\boxed{0.99997} \)