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Allushta [10]
3 years ago
11

Julia performs an experiment to measure the wavelength of four different waves and records her data in the table below.

Physics
2 answers:
Nitella [24]3 years ago
7 0

Explanation :

The amplitude of the wave illustrates the energy of the wave. The energy of the depends on the amplitude and the frequency of the wave.

The maximum distance covered by the wave is called its amplitude. The topmost point on the wave is called its crest and the down most point is called its trough (as shown in the attached figure).

From the given data, the wave having the amplitude of 3 cm has the least energy while the wave having an amplitude of 12 cm have maximum energy.

So, the correct order of the wave from the lowest energy to the highest energy is :

1 < 4 < 2 < 3

Hence, this is the required solution.

Svet_ta [14]3 years ago
4 0
The correct order they go in is "1-4-2-3"  The correct answer is D.
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The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.
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Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}

Here, R_{\infty} is the "general" Rydberg constant, m_e is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, M=1.67*10^{-27}kg:

R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}

Now, we calculate the wavelength for hydrogen:

\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm

For deuterium, we have M=2(1.67*10^{-27}kg):

R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm

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3 years ago
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