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Softa [21]
2 years ago
6

Calculate the wavelength of the photon emitted when an electron makes a transition from n=6 to n=3. You can make use of the foll

owing constants: h=6.626×10−34 J⋅s c=2.998×108 m/s 1 m=109 nm
Chemistry
1 answer:
Angelina_Jolie [31]2 years ago
6 0

<u>Answer:</u> The wavelength of light is 1.094\times 10^{-6}m

<u>Explanation:</u>

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

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

Where,

\lambda = Wavelength of radiation

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

n_f = Final energy level = 3

n_i = Initial energy level = 6

Putting the values in above equation, we get:

\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{6^2} \right )\\\\\lambda =\frac{1}{914617m^{-1}}=1.094\times 10^{-6}m

Hence, the wavelength of light is 1.094\times 10^{-6}m

You might be interested in
For the reaction 2N2O5(g) &lt;---&gt; 4NO2(g) + O2(g), the following data were colected:
KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0
2 years ago
Determine the number of neutrons in an atom of Rg-272.
IRISSAK [1]
In order to find the number of neutrons in the atom,

you need to calculate the difference between the top and bottom numbers

which means 272 - 111 = 161

Hope this helps
6 0
2 years ago
A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?
Tanzania [10]
<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume =  0.50 cm × 10.0 cm × 2.0 cm

             = 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

                                      = 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0
2 years ago
The recommended daily intake of potassium (K) is 4.725 grams. Assuming that every raisin contains 3.677 milligrams of K, how man
Pani-rosa [81]
3.677mg =0,003677g \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ 1 \ raisin\\ 4,725g \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ x\\\\ x=\frac{4,725g*1}{ 0,003677g}\approx 1285 \ raisins
3 0
2 years ago
Classify the following as a type of potential energy or kinetic energy (use the letters K or P)
Zanzabum
K, P, K, K, P, K, K, P, K, P. If it is moving, it is kinetic, if it isn't, it's potential. the sugar one is a little tricky using that method though, because we generally consider this in terms of spacial movement, but sugar holds energy which is later released by your body to allow you to move.the chemical bonds have potential energy because they release energy when broken.
5 0
2 years ago
Read 2 more answers
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