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3 years ago

13

When an electron in a certain excited energy level in a one-dimensional box of length 2.00 Å makes a transition to the ground st

ate, a photon of wavelength 8.79 nm is emitted. Find the quantum number of the initial state?

1 answer:

Fiesta28 [93]3 years ago

5 0

Answer:

Calculate the wavelength associated with an electron with energy 2000 eV.

Sol: E = 2000 eV = 2000 × 1.6 × 10–19 J

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The main illustration in the video shows the life track of a one-solar mass star. Each point along this track represents _______

REY [17]

Each point along the track of one solar mass star represents the star's surface temperature and luminosity at one time.

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A star having a mass equal to the mass of the Sun is called a one-solar mass star.

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1 year ago

What is the speed of a wave with a frequency of 6 Hz and a wavelength of 12 m? No links

myrzilka [38]

72 m/s

Explanation:

Given,

Frequency ( f ) = 6 Hz

Wavelength ( λ ) = 12 m

To find : -

Speed ( v ) = ?

Formula : -

v = f x λ

v

= 6 x 12

= 72 m/s

Therefore,

the speed of a wave with a frequency of 6 Hz and a wavelength of 12 m is 72 m/s.

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2 years ago

Nitrogen fixation is the conversion of

Varvara68 [4.7K]

Answer:

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6 0

2 years ago

When electrons are lost, a blank ion is formed

Serhud [2]

Answer:

Positive ions, or cations.

3 0

3 years ago

n the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed o

Ivahew [28]

Answer:

(a) $T=1.5*10^{-6}s$

(b) $I=1.1*10^{-3}A$

(c) $\mu=9.71*10^{-24}A\cdot m^2$

Explanation:

(a) The orbital period is the time that the electron spend to travel the orbit of the atom. Thus, it is given by the length of the circular orbit divided by its velocity:

$T=\frac{2\pi r}{v}\\T=\frac{2\pi(5.3*10^{-11}m)}{2.2*10^{6}\frac{m}{s}}\\T=1.5*10^{-6}s$

(b) Current means charge over time, So, in this case is charge over period:

$I=\frac{q}{t}\\I=\frac{e}{T}\\I=\frac{1.6*10^{-19}C}{1.5*10^{-6}s}\\\\I=1.1*10^{-3}A$

(c) Magnetic moment is given by:

$\mu=IA$

Here A is the area of the orbit.

$\mu=I\pi r^2\\\mu=(1.1*10^{-3}A)\pi(5.3*10^{-11}m)^2\\\mu=9.71*10^{-24}A\cdot m^2$

4 0

3 years ago