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dalvyx [7]
3 years ago
5

Determine the wavelength of incident electromagnetic radiation required to cause an electron transition from the n = 5 to the n

= 7 level in a hydrogen atom.
Physics
1 answer:
ollegr [7]3 years ago
8 0
The correct answer is: wavelength = 4562 nm

Explanation:

Rydberg's formula is given as:
\frac{1}{\lambda} = R[ \frac{1}{n_1^2}  - \frac{1}{n_2^2} ] --- (1)

Where 
R = Rydberg's constant = 1.096 * 10^7 per meter
n_1 = 5
n_2 = 7

λ = Wavelength

Plug in the values in (1):

(1)=> \frac{1}{\lambda} = (1.096 * 10^7)[ \frac{1}{5^2} - \frac{1}{7^2} ]

\frac{1}{\lambda} = (1.096 * 10^7)[ 0.04 - 0.020 ] \\ \lambda =  \frac{1}{(1.096 * 10^7)[0.020 ]} \\ \lambda = 4562 nm
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What characteristics determine how easily two substances change temperature
ValentinkaMS [17]

Answer;

Amount of time the two substances are in contact

Area in contact between the two substances

Specific heat of the material that makes up the substances

Explanation;

The change in temperature of a substance is caused by heat energy. The change in temperature will depend on factors such as mass of the substance, the type of material it is made from, the time taken , specific heat of the material that makes the substance, and also the area of contact.

The amount of time the two substances are in contact affect the change in temperature such that if the two bodies are in contact for a longer time then a bigger change in temperature will be observed.

Specific heat capacity also determines the change in temperature that will be observed, such that a substance with a bigger specific heat capacity will record a small change in temperature.

7 0
3 years ago
What is the density of an object that has a mass of 30 g and a volume of 20cm cubed/ to the third power?
Ksenya-84 [330]

Answer:

d= 1.5 g/cm3

Explanation:

datos

m= 30g

v= 20cm3

d=?

formula

d= m / v

solución

d= 30g / 20cm3 = 1.5g/cm3

6 0
2 years ago
Q1. Which statement is correct
marishachu [46]

Answer:

C

D

Explanation:

7 0
2 years ago
A car of mass 750 kg accelerates away from traffic lights. At the end of the first 100 m it has reached a speed
malfutka [58]

the work done on the car by the force of its engine is 78,000 J.

" The work done on the car by the force of friction is 24,000 J.

Increasing the car's kinetic energy at the end of the first 100 m is 54,000J

a. Completed work = force x distance. Engine output = 780 N, that is,

780 N x 100 m = 78,000 J.

b. Completed work = force x distance. Friction force = 240 N, that is,

240 N x 100 m = 24,000 J.

c. Kinetic energy = 1 \ 2 x m x v2

= 1 \ 2 x 750 kg x 12 squared = 375 x 144 = 54,000 J.

<h3>How powerful is the engine of a car? </h3>

Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.

Learn more about work done here:  brainly.com/question/25573309

#SPJ10

5 0
2 years ago
A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
svp [43]

a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

\omega=\sqrt{\frac{k}{m}} is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s

The amplitude, A, can be found from the maximum velocity of the spring:

v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m

So, the equation of motion is

x(t)= 2.0 sin (10 t) [m]

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

U(x)=\frac{1}{2}kx^2

While the kinetic energy is given by:

K=\frac{1}{2}mv^2

The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

However, \frac{m}{k}=\frac{1}{\omega^2}, so we have

(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

with two solutions:

\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum. By using T=0.628 s, we find

L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






5 0
3 years ago
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