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soldi70 [24.7K]
3 years ago
6

Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest

wavelength in the series.
Physics
1 answer:
mr_godi [17]3 years ago
8 0

Answer and Explanation:

The computation of the shortest wavelength in the series is shown below:-

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

Where

\lambda represents wavelength

R represents Rydberg's constant

n_f represents Final energy states

and n_i represents initial energy states

Now Substitute is

1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )

now we will put the values into the above formula

= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )

= 1218888.889 m^{-1}

Now we will rewrite the answer in the term of \lambda

\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m

So, the whole Paschen series is in the part of the spectrum.

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Answer:

Frequency = 1,550Hz

Explanation:

To solve this we can use the equation: f=\frac{v}{\lambda}

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We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m

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(Which I believe is just below a G6 if you were interested)

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4 0
3 years ago
In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.
vaieri [72.5K]

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X

^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

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^{211}_{83}Bi\Rightarrow ^{211}_{84}Po

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(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

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^{22}_{11}Na\Rightarrow ^{22}_{10}Ne

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

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