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Natasha_Volkova [10]
3 years ago
9

An excited hydrogen atom emits light with a wavelength of 486.4 nm to reach the energy level for which n = 2. In which principal

quantum level did the electron begin?
Physics
1 answer:
SpyIntel [72]3 years ago
5 0

Answer:

The electron began in the quantum level of 4

Explanation:

Using the formula of wave number:

Wave Number = 1/λ = Rh(1/n1² - 1/n2²)

where,

Rh = Rhydberg's Constant = 1.09677 x 10^7 /m

λ = wavelength of light emitted = 486.4 nm = 486.4 x 10^-9 m

n1 = final shell = 2

n2 = initial shell = ?

Therefore,

1/486.4 x 10^-9 m = (1.09677 x 10^7 /m)(1/2² - 1/n2²)

1/4 - 1/(486.4 x 10^-9 m)(1.09677 x 10^7 /m) = 1/n2²

1/n2² = 0.06082

n2² = 16.44

<u>n2 = 4</u>

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The weight of an object on the surface of mass is 1850 Newton and its mass is 500kg.Find the acceleration due to gravity on mars
Fudgin [204]

F = m.a

a = F/m

a = 1850/500

a = 3.7 m/s²

3 0
2 years ago
What causes the bright lines seen in the emission spectrum?
Snezhnost [94]
Base in your questions that ask what cause the bright lines seen in the emission spectrum and i think the best answer to that is the H2 gas is used when protons was heated so the electron absorb all the photons and get exited and resulted by given of a light.
7 0
3 years ago
A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

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

Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

Learn more about Newton's laws of motion and accelerated motion:

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6 0
4 years ago
3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage dropped across each phase?
ZanzabumX [31]

Answer:

E_s = 277.13V

Explanation:

Given

Load\ Voltage = 480V

Required

Determine the voltage dropped in each stage.

The relation between the load voltage and the voltage dropped in each stage is

E_l = E_s * \sqrt3

Where

E_l  = 480

So, we have:

480 = E_s * \sqrt3

Solve for E_s

E_s = \frac{480}{\sqrt3}

E_s = \frac{480}{1.73205080757}

E_s = 277.128129211

E_s = 277.13V

<em>Hence;</em>

<em>The voltage dropped at each phase is approximately 277.13V</em>

5 0
3 years ago
A weather balloon is floating at a constant height above the earth when it releases a pack of instruments. If the pack hits the
andrezito [222]

Answer:

c 275 m

Explanation:

Given parameters:

Final velocity  = 73.5m/s

Unknown:

Height of fall  = ?

Solution:

Since the body is falling from rest, U = 0 or initial velocity is 0m/s. Then we use one of the kinematics equation to solve this problem.

           V²   = U²  + 2gH

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height

           73.5²   = 0²  +  (2 x 9.8 x h)

         5402.25 = 19.6h

                h = 275.6m

4 0
3 years ago
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