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const2013 [10]
3 years ago
7

Consider the transition from the energy levels n = 3 to n = 5. What is the wavelength associated with this transition, in nm?

Physics
2 answers:
slavikrds [6]3 years ago
8 0

<u>Answer:</u> The wavelength of transition is 1281 nm

<u>Explanation:</u>

To calculate the wavelength of light, we use Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 1.097\times 10^7m^{-1}

n_f = Higher energy level = 5

n_i= Lower energy level = 3

Putting the values in above equation, we get:

\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{5^2} \right )\\\\\lambda =\frac{9\times 25}{0.2518\times 10^7m^{-1}\times 16}=12.81\times 10^{-7}m

Converting this into nanometers, we use the conversion factor:

1m=10^9nm

So, 12.81\times 10^{-7}m\times (\frac{10^9nm}{1m})=1281nm

Hence, the wavelength of light is 1281 nm

s344n2d4d5 [400]3 years ago
5 0

Answer:

wavelength is 1301.8 nm

Explanation:

Given data

energy levels n = 3 to n = 5

to find out

What is the wavelength

solution

we know wavelength formula that is given below

1/wavelength = RH(1/n1² - 1/n2²)

here we know RH value is  10973731.6 m^(-1)

so

wavelength = 1 / RH(1/n1² - 1/n2²)

wavelength = 1 / 10973731.6 (1/3² - 1/5²)

wavelength = 1.3018 ×10^-6 m

wavelength is 1301.8 nm

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2 years ago
2. A wave on a rope has a wavelength of 2.0 m and a frequency of 2.0 Hz. What is the speed of the
TEA [102]

Answer:

4 m/s or 4 meters per second.

Explanation:

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7 0
3 years ago
A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against
valkas [14]

Answer:

(A) Work done will be 87.992 KJ

(B) Work done will be 167.4 KJ            

Explanation:

We have given mass of methane m = 4.5 gram = 0.0045 kg

Volume occupies V_1=12.7dm^3=12.7liters

And volume is increased by 3.3dm^3 so V_2=12.7+3.3=16liters

Temperature T = 310 K

Pressure is given as 200 Torr = 26664.5 Pa

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W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj

(b) At reversible process work done is given by W=nRTln\frac{V_2}{V_1}

We have given mass = 4.5 gram

Molar mass of methane = 16

So number of moles n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125

So work done W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J

7 0
3 years ago
A 1600kg cannon fires a 5kg cannonball horizontally. The exit velocity of the cannonball is 80m/s and the barrel length is 2m. W
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Answer:

 a = 1600 m / s²

Explanation:

For this exercise we use the kinematics relations,

         v² = v₀² + 2 a x

where v₀ is the initial velocity of the bullet, which as part of rest is zero, for the distance (x) we can assume that the gases accelerate along the entire trajectory of the cannon x = 2m

         a = \frac{v^2}{2x}

let's calculate

         a = \frac{80^2}{2 \ 2}

         a = 1600 m / s²

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True or False: The average atomic mass is always closer to the isotope with the largest mass
svlad2 [7]
The awnser would be True. Hope I helped!
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3 years ago
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