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3 years ago

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Find the frequency of light emitted in the transition from the 148th orbit to the 139th orbit. Find the frequency of light absor

bed in the transition from the 165th orbit to the 174th orbit. Submit Answer

1 answer:

Romashka-Z-Leto [24]3 years ago

8 0

Answer:

$f=3\times 10^8\times 64.90=19.4\times 10^9 s^{-1}$

$f=3\times 10^8\times39.9537=119.8613\times 10^8\ /sec$

Explanation:

for the transition from 148th orbit to 138th orbit

we know the formula

$\frac{1}{\lambda }=R\times \left ( \frac{1}{n^2_{final}}-\frac{1}{n^2_{initial}} \right )$

where R is Rydberg constant whose value is $1.0974\times 10^7 /m$

putting this value in equation

$\frac{1}{\lambda }=1.0974\times 10^7 \left ( \frac{1}{139^2}-\frac{1}{148^2} \right )$

$\frac{1}{\lambda }=64.9075$

$f=c\lambda$

$f=3\times 10^8\times 64.90=19.4\times 10^9 s^{-1}$

for the transition from 165th to 174th orbit

we know the formula

$\frac{1}{\lambda }=R\times \left ( \frac{1}{n^2_{final}}-\frac{1}{n^2_{initial}} \right )$

where R is Rydberg constant whose value is $1.0974\times 10^7 /m$

putting this value in equation

$\frac{1}{\lambda }=1.0974\times 10^7 \left ( \frac{1}{165^2}-\frac{1}{174^2} \right )$

$\frac{1}{\lambda }=39.95377$

$f=c\lambda$

$f=3\times 10^8\times39.9537=119.8613\times 10^8\ /sec$

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They are both created by waves of different forms of energy... sound is the oscillation of other substances, called a medium, while the electromagnetic waves are oscillating through electromagnetic energy.

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4 years ago

A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the

MissTica

Answer : The value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

$q=n\times c_v\times \Delta T$

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

$\Delta T$ = Change in temperature = 15.7 K

$c_v$ = heat capacity at constant volume of $N_2$ (diatomic molecule) = $\frac{5}{2}R$

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

$q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K$

$q=286.2J$

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

As we know that, work done is zero at constant volume. So,

$\Delta U=q=286.2J$

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4 years ago

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Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th

Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

$\theta=30^{\circ}$

So, the ball's horizontal velocity is

$v_x = u cos \theta = (15)(cos 30)=13.0 m/s$

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

$y=u_yt+\frac{1}{2}at^2$

where

$u_y = u sin \theta = (15)(sin 30) = 7.5 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

$2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0$

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

$d=v_x t =(13.0)(1.19)=15.5 m$

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

$v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s$

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

$d=98 m$

while the horizontal velocity of the ball is

$v_x = u cos \theta = (34)(cos 30)=29.4 m/s$

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

$t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s$

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

$y= h + u_y t + \frac{1}{2}at^2$

where

h = 1.2 m is the initial height

$u_y = u sin \theta = (34)(sin 30)=17.0 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting t = 3.33 s,

$y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m$

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

$u_y' = 7 m/s$

So its position of the glove at time t' is

$y'= h' + u_y' t' + \frac{1}{2}at'^2$

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

$3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0$

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

$t'' = t -t'$

So we have two solutions:

$t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s$

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

I need help please...

Natalka [10]

No cluuuueee :/ sowwwwyyy but good luck

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3 years ago

Read 2 more answers

if a body of mass 2kg travels with 2 m/s and other body of mass 2kg travels with 1m/s then find v1 and v2 in given statement

KIM [24]

Answer:

Your answer is here,

Explanation:

v1 = 4 kg m/s

v2 = 2 kg m/s

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