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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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3 years ago
Differentiate between rest and motion.
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<em>1</em><em>. </em><em>A body is said to be at rest if its position does not change with respect to its surroundings.</em>

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Why can’t solids flow
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3 years ago
Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show
Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

m_2 = \frac{20}{3} kg

Part b)

Angular acceleration is given as

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = 176.6 N

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

m_1g r_1 = m_2 g r_2

20 g(0.5) = m_2 g(1.5)

10 = 1.5 m_2

m_2 = \frac{20}{3} kg

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

m_1g - T = m_1 a

also we have

T r_1 = I\alpha

now we have

m_1g = \frac{I a}{r_1^2} + m_1 a

a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}

a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}

a = 0.981 m/s^2

so angular acceleration is given as

\alpha = \frac{a}{r_1}

\alpha = \frac{0.981}{0.5}

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = \frac{I\alpha}{r_1}

T = \frac{45 (1.96)}{0.5}

T = 176.6 N

7 0
3 years ago
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