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valentina_108 [34]
2 years ago
14

What is the value of n in the balmer series for which the wavelength is 410.2 nm.?

Physics
1 answer:
m_a_m_a [10]2 years ago
3 0

The answer is n= 6.

What is Balmer series?

The Balmer series is the portion of the emission spectrum of hydrogen that represents electron transitions from energy levels n > 2 to n = 2. These are four lines in the visible spectrum. They are also known as the Balmer lines. The four visible Balmer lines of hydrogen appear at 410 nm, 434 nm, 486 nm and 656 nm.

For the Balmer series, the final energy level is always n=2. So, the wavelengths 653.6, 486.1, 434.0, and 410.2 nm correspond to n=3, n=4, n=5, and n=6 respectively. Since the last wavelength, 410.2 nm, corresponds to n=6, the next wavelength should logically correspond to n=7.

To solve for the wavelength, calculate the individual energies, E2 and E7, using E=-hR/(n^2). Then, calculate the energy difference between E2 (which is the final) and E7 (which is the initial). Finally, use lamba=hc/E to get the wavelength.

To learn more about emission spectrum click on the link below:

brainly.com/question/24213957

#SPJ4

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a rocket engine can accelerate a rocket launched from rest vertically up with an acceleration of 22.9 m/s2. however, after 50.0
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The acceleration of a rocket engine is given here, and after 50 seconds of flight, the engine fails, and we must determine the altitude of the rocket at the time the engine fails.  Because the rocket starts from rest, the time taken is 50 seconds, the initial velocity is zero, and the acceleration is 22.9 meters per second square. So we use the kinamatics equation s equal to v. I t plus half 8 square. There is no acceleration at the start. 22.9 and t is 50 seconds, so displacement 2.86 times 10 to the power 4 is met. This is the rocket's displacement in 50 seconds, so this is the rocket's altitude when the engine fails.

<h3>What exactly is accelerate?</h3>
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Hence, There is no acceleration at the start. 22. This is the rocket's displacement in 50 seconds, so this is the rocket's altitude when the engine fails.

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3 0
1 year ago
the potential at a point A is 30 V higher than at point B. How much work would you do moving 4.0 C charge from B to A?
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7 0
3 years ago
A 5.75 × 107 kg battle ship originally at rest fires a 1100-kg artillery shell horizontally with a velocity of 550 m/s. show ans
prohojiy [21]

Answer:

-0.01052 m/s

Explanation:

M = mass of ship = 5.75\times 10^7\ kg

m = mass of shell = 1100 kg

v = velocity of shell = 550 m/s

u = recoil velocity of ship

As linear momentum is conserved

(M - m)u=-mv\\\Rightarrow u=-\frac{mv}{M - m}\\\Rightarrow u=-\frac{1100\times 550}{5.75\times 10^7+1100}\\\Rightarrow u=-0.01052\ m/s

The recoil velocity of the ship taking the firing direction to be the positive direction is -0.01052 m/s

6 0
3 years ago
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