MATHEMATICS COLLEGE

Determine the wavelengths of all the possible photons that can be emitted from the n = 5 state of a hydrogen atom.

Answers

Answer 1

Answer:

Wavelengths of all possible photons are;

λ1 = 9.492 × 10^(-8) m

λ2 = 1.28 × 10^(-6) m

λ3 = 1.28 × 10^(-6) m

λ4 = 4.04 × 10^(-6) m

Step-by-step explanation:

We can calculate the wavelength of all the possible photons emitted by the electron during this transition using Rydeberg's equation.

It's given by;

1/λ = R(1/(n_f)² - 1/(n_i)²)

Where;

λ is wavelength

R is Rydberg's constant = 1.0974 × 10^(7) /m

n_f is the final energy level = 1,2,3,4

n_i is the initial energy level = 5

At n_f = 1,.we have;

1/λ = (1.0974 × 10^(7))(1/(1)² - 1/(5)²)

1/λ = 10535040

λ = 1/10535040

λ = 9.492 × 10^(-8) m

At n_f = 2,.we have;

1/λ = (1.0974 × 10^(7))(1/(2)² - 1/(5)²)

1/λ = (1.0974 × 10^(7))(0.21)

1/λ = 2304540

λ = 1/2304540

λ = 4.34 × 10^(-7) m

At n_f = 3, we have;

1/λ = (1.0974 × 10^(7))(1/(3)² - 1/(5)²)

1/λ = (1.0974 × 10^(7))(0.07111)

1/λ = 780373.3333333334

λ = 1/780373.3333333334

λ = 1.28 × 10^(-6) m

At n_f = 4, we have;

1/λ = (1.0974 × 10^(7))(1/(4)² - 1/(5)²)

1/λ = (1.0974 × 10^(7))(0.0225)

1/λ = 246915

λ = 1/246915

λ = 4.04 × 10^(-6) m

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Answer:

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The answer to this it d.

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Step-by-step explanation:

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Peter baked 64 loaves of bread in three days. How many loaves did he bake each day, if he baked 3 more loaves on the second day than on the first day, and 4 more loaves on the third day than on the first day?

Answers

Answer:

Step-by-step explanation:

1st day: x

2nd day: x+3

3rd day: x+4

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You have a triangle that has side lengths of 6,9,and 12. Give the side lengths of a similar triangle that is smaller than the given triangle.

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Those lengths have a common factor of 3. Removing that factor gives you the smaller similar triangle with sides 2, 3, and 4.

What's the square root of 25, 100, 36, 84, and 4.

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Step-by-step explanation:

$√(25) = \pm \: 5 \n \n √(100) = \pm \: 10 \n \n √(36) = \pm \: 6 \n \n √(84) = \pm \: 9.165\n \n √(4) = \pm \: 2 \n \n$

Compute the area of the region D bounded by xy=1, xy=16, xy2=1, xy2=36 in the first quadrant of the xy-plane. Using the non-linear change of variables u=xy and v=xy2, find x and y as functions of u and v.x=x(u,v)= ?

y=y(u,v)=?

Find the determinant of the Jacobian for this change of variables.

∣∣∣∂(x,y)/∂(u,v)∣∣∣=det=?

Using the change of variables, set up a double integral for calculating the area of the region D.

∫∫Ddxdy=?

Evaluate the double integral and compute the area of the region D.

Area =

Answers

Answer:

53.7528

Step-by-step explanation:

Notice that when

$xy = 1 ,\,\,\, xy = 16 , \,\,\, xy^2 = 1 \,\,\,, xy^2 = 36 \n\n$

If you set

$u = xy , v = xy^2$

as they suggest, then

$y = (v)/(u)} \,\,\,\, \text{and} \,\,\,\, \n\n{\displaystyle x = (u)/(y) = (u)/(v/u) = (u^2)/(v)$

Then

${\diplaystyle (\partial(x,y))/(\partial(u,v))} =\det \begin{pmatrix} 2u/v && -u^2/v^2 \n -v/u^2 && 1/u \end{pmatrix} = (1)/(v) }$

Therefore

$\iint\limits_(D) dx\,dy = \int\limits_(1)^(36)\int\limits_(1)^(16) (1)/(v) \, du \, dv = 15 \ln(36) = 53.7528$

A Jacobian matrix is formed by the first partial derivatives of a multivariate function that utilizes a training algorithm, and further calculation as follows:

Jacobian:

To evaluate the integral, cover the bounds, the integrand, and the differential area dA.

Transform the four equations in terms of u and v, notice that $u= xy \ \ and \ \ xy = 1, xy = 16$

implies that $1\leq u \leq 16.$

Similarly, $v= xy^2\ \ and\ \ xy^2= 1 , xy^2= 25$ implies that $1 \leq v \leq 25$

so write this integration region as $S= {(u,v) |1 \leq u \leq 18, 1 \leq v \leq 25}.$

Translate the equations from uv - plane to xy- plane. It is obtained by solving,

$u= xy, y= xy^2 \n\n\left.\begin{matrix}u=xy & \n v=xy^2& \end{matrix}\right\} \to \left.\begin{matrix}u^2=x^2y^2 & \n v=xy^2& \end{matrix}\right\} \n\n\to x=(u^2)/(v), y=(v)/(u)$

Convert dA part of the integral , using is $dA= |(\partial (x,y))/(\partial(u,v))| dudv.$

That is, $dA= \begin{vmatrix}(\partial x)/(\partial u) & (\partial x)/(\partial v)\n (\partial y)/(\partial u) & (\partial y)/(\partial v) \end{vmatrix} \ du dv \n\n$

Sampule the partial derivatives to find the Jacobian.

$dA=\begin{vmatrix}(2u)/(v) &-(u^2)/(v^2) \n -(v)/(u^2) &(1)/(u) \end{vmatrix} \ dudv\n\n=[((2u)/(v)) ((1)/(u)) -(- (u^2)/(v^2))(-(v)/(u^2))]\ du dv\n\n=((2)/(v)- (1)/(v)) \ dudv\n\n=(1)/(v)\ du dv\n\n$

The Jacobian the transformation is $dA= (1)/(v)dudv$

The region is $S={(u,v) |1\leq u \leq 16, 1\leq v\leq 25}.$

Rewrite the integral, using the transformation: $S,\ x=(u^2)/(v) =, y=(v)/(u) \ \ and\ \ dA=(1)/(v) dudv\n\n\int\int_R 1dA =\int \int_S (1)/(v)\ dudv= \int^(25)_(1) \int^(16)_(1) \ (1)/(v) \ dudv\n\n$

Evaluate the inner integral with respect to u.

$\to \int\int_R 1dA = \int^(25)_(1) \int^(16)_(1) \ (1)/(v) \ dudv\n\n$

by solving the value we get

$= 30 \ ln (5) \approx 48.28$

Find out more about the Jacobians here:

brainly.com/question/9381576

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