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2 years ago

Please I need help! This is the last question I need for this assignment!

Physics

1 answer:

Ne4ueva [31]2 years ago

6 0

Answer:

When the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.

What is specific heat capacity?

Specific heat capacity is the quantity

of heat required to raise a unit mass of

a substance by 1 kelvin.

Specific heat capacity of water and sand

{<em>refer to the above attachment}</em>

Δθ = Q/mc

Thus, for an equal mass of water and sand, when the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of n is $n =7$

Explanation:

From the question we are told that

The value of m = 2

For every value of $m, n = m+ 1, m+2,m+3,....$

The modified version of Balmer's formula is $\frac{1}{\lambda} = R [\frac{1}{m^2} - \frac{1}{n^2} ]$

The Rydberg constant has a value of $R = 1.097 *10^{7} m^{-1}$

The objective of this solution is to obtain the value of n for which the wavelength of the Balmer series line is smaller than 400nm

For m = 2 and n =3

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{3^2} ]$

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$\lambda = 656nm$

For m = 2 and n = 4

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{4^2} ]$

$\lambda = \frac{1}{2056875}$

$\lambda = 486nm$

For m = 2 and n = 5

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{5^2} ]$

$\lambda = \frac{1}{2303700}$

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For m = 2 and n = 6

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{6^2} ]$

$\lambda = \frac{1}{2422222}$

$\lambda = 410nm$

For m = 2 and n = 7

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{7^2} ]$

$\lambda = \frac{1}{2518622}$

$\lambda = 397nm$

So the value of n is 7

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3 years ago

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4 years ago

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2 years ago

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Answer:

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$\Delta U=4620 J$ is the gain in potential energy

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3 years ago

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