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2 years ago

Using the rydberg constant determined in question 1, calculate the shortest wavelength in the paschen series

Physics

1 answer:

salantis [7]2 years ago

7 0

The shortest wavelength in the paschen series= $8.2\times 10^{-7}$ m.

<h3>How do we calculate the shortest wavelength in the paschen series?</h3>

Emission lines for hydrogen occur when electrons drop from some energy level to a lower energy level. To calculate the shortest wavelength in the paschen series we are using the formula,

$\frac{1}{\lambda} =R_{H} (\frac{1}{n_{f}^{2} }-\frac{1}{n^{2} } )$

Here, we are given,

$R_{H}$ = Rydberg constant= $1.09737 \times 10^{7} m^{-1}$

$n_f$ = The lower energy level quantum number.=3 (for the paschen series).

n= The quantum number of whichever state the transitions occur from = (for this case of the paschen series).

We have to find the wavelength associated with the photon emitted = $\lambda$ m.

Now we substitute the known values in the above equation, we can find that,

$\frac{1}{\lambda} =1.09737 \times 10^{7} (\frac{1}{3^2 }-\frac{1}{\infty^{2} } )$

Or, $\frac{1}{\lambda} =1.09737 \times 10^{7}\times \frac{1}{9 }$

Or, $\frac{1}{\lambda} =1,219,300$

Or, $\lambda= 8.2\times 10^{-7}$ m

From the above calculation we can conclude that the shortest wavelength in the paschen series is $8.2\times 10^{-7}$ m

Learn more about paschen series:

brainly.com/question/15322810

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Which of the following expressions gives the ratio of the energy density of the magnetic field to that of the electric field jus

miss Akunina [59]

Answer:

$(d) \ \ \frac{\mu_o}{\epsilon_o} (\frac{L}{2\pi r*R} )^2$

Explanation:

Energy density in magnetic field is given as;

$U_B = \frac{1}{2 \mu_o} B^2$

where;

B is the magnetic field strength

Energy density of electric field

$U_E = \frac{1}{2}\epsilon E^2$

where;

E is electric field strength

Take the ratio of the two fields energy density

$\frac{U_B}{U_E} = \frac{1}{2\mu_o} B^2 / \frac{1}{2}\epsilon E^2\\\\\frac{U_B}{U_E} = \frac{B^2}{2\mu_o} *\frac{2}{\epsilon E^2} \\\\\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B^2}{E^2})$

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B}{E})^2$

But, Electric field potential, V = E x L = IR (I is current and R is resistance)

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{E*L})^2$

Now replace E x L with IR

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{IR})^2$

Also, B = μ₀I / 2πr, substitute this value in the above equation

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{\mu_oI*L}{2\pi r* IR})^2$

cancel out the current "I" and factor out μ₀

$\frac{U_B}{U_E} = \frac{\mu_o^2}{\mu_o \epsilon} (\frac{L}{2\pi r* R})^2$

Finally, the equation becomes;

$\frac{U_B}{U_E} = \frac{\mu_o}{\epsilon} (\frac{L}{2\pi r*R })^2$

Therefore, the correct option is (d) μ₀/ϵ₀ (L /R 2πr)²

3 0

3 years ago

If earth increase the distance from the sun, what will happen to the period of orbi t(the time it takes to complete one revoluti

Mandarinka [93]

The period of the orbit would increase as well

Explanation:

We can answer this question by applying Kepler's third law, which states that:

"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"

Mathematically,

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Where

T is the orbital period

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