We use the Rydberg Equation for this which is expressed as:

1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]

where lambda is the wavelength, where n represents the final and initial states. Brackett series means that the initial orbit that electron was there is 4 and R is equal to 1.0979x10^7m. Thus,

1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
1/1.0979x10^7m = 1.0979x10^7m [ 1/(n2)^2 - 1/(4)^2]

Solving for n2, we obtain n=1.

5 0

3 years ago

If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×1

Ipatiy [6.2K]

This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

$I=\frac{Q}{t}$

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

$\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}$

As you can observe above, the division of the powers was solved by just subtracting their exponents.

<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>

3 0

1 year ago