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Nutka1998 [239]

2 years ago

8

An electron in a hydrogen atom undergoes a transition from the n = 3 level to the n = 6 level. To accomplish this, energy, in th

e form of light, must be absorbed by the hydrogen atom. Calculate the energy of the light (in kJ/photon) associated with this transition.

1 answer:

Harlamova29_29 [7]2 years ago

7 0

Answer: $1.816\times 10^{-19} J$

Explanation:

Given

$E=\frac{hc}{\lambda }$

$E=2.18\times 10^{-18}(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

where h=Planck constant

c=speed of light

$E=2.18\times 10^{-18}(\frac{1}{3^2}-\frac{1}{6^2})$

$E=2.18\times 10^{-18}\times \frac{1}{12}$

$E=1.816\times 10^{-19} J$

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Find the instantaneous acceleration at t=ls for an object moving along a straight axis with velocity function:

Maru [420]

The answer is A.
Explanation:
We know that the average acceleration a for an interval of time Δt is expressed as:

a = Δv
Δt
where Δv is the change in velocity that occurs during Δt.
e formula for the instantaneous acceleration a is almost the same, except that we need to indicate that we're interested in knowing what the ratio of Δv to Δt approaches as Δt approaches zero.

We can indicate that by using the limit notation.

So, the formula for the instantaneous acceleration is:

a = lim Δv
Δt→0 Δt

8 0

2 years ago

Calculate the gravitational potential energy of a 2 kg banana hanging 5

Hunter-Best [27]

Answer:

100 J

Explanation:

The potential energy is given by the formula ...

PE = mgh

= (2 kg)(10 m/s^2)(5 m) = 100 J

7 0

3 years ago

A child pushes a toy box across the floor in 5 seconds. If he did 30 J of work on the toy box, what amount of power was required

s2008m [1.1K]

My answer is 6 watts because 30J/5s is 6

4 0

2 years ago

Read 2 more answers

True or false your vocal cords act like a guitar string when it comes to sound

torisob [31]

Answer:

True

Explanation:

8 0

2 years ago

Read 2 more answers

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

2 years ago