A jet airplane has a velocity of 1145 knots. A knot is 1 nautical mile (nm)/hr. A nautical

Calculate the wavelength of the photon that would be absorbed or emitted. Round your answer to 3 significant digits.

Flauer [41]

This is an incomplete question, the image for the given question is attached below.

Answer : The wavelength of photon would be absorbed, $3.06\times 10^{-7}m$

Explanation :

From the given diagram of energy we conclude that,

Energy at ground state, A = 400 zJ

Energy of 2nd excited state, C = 1050 zJ

Now we have to calculate the energy of the photon.

$E=E_c-E_A$

$E=(1050-400)zJ= 650zJ=650\times 10^{-21}J$

Now we have to calculate the wavelength of the photon.

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

E = energy of photon = $650\times 10^{-21}J$

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength of photon = ?

c = speed of light = $3\times 10^8m/s$

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

$650\times 10^{-21}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}$

$\lambda=3.06\times 10^{-7}m$

Therefore, the wavelength of photon would be absorbed, $3.06\times 10^{-7}m$

5 0

3 years ago

In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53 × x10−

sergiy2304 [10]

the force between the electron and the proton.
a) Use F = k * q1 * q2 / d²
where k = 8.99e9 N·m²/C²
and q1 = -1.602e-19 C (electron)
and q2 = 1.602e-19 C (proton)
and d = distance between point charges = 0.53e-10 m
The negative result indicates "attraction".

the radial acceleration of the electron.
b) Here, just use F = ma
where F was found above, and
m = mass of electron = 9.11e-31kg, if memory serves
a = radial acceleration

the speed of the electron.
c) Now use a = v² / r
where a was found above
and r was given

<span> the period of the circular motion.</span>
d) period T = 2π / ω = 2πr / v
where v was found above
and r was given

3 0

3 years ago

Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How ma

Katen [24]

Answer: 12.0 milliliters of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn.

Explanation:

moles = $\frac{\text {given mass}}{\text {Molar mass}}$