Lines of a particular element appear at the same wavelength in both emission and absorption line spectra. True or False

Physics

1 answer:

Liula [17]4 years ago

3 0

<h2>Answer: True </h2><h2></h2>

Each atom emits or absors electromagnetic radiation, but only in some frequencies that are characteristic of its chemical element.

Now, if we supply energy in the form of heat (thermal energy) to a certain element in its gas phase, this will be stimulated and its atoms will emit radiation in certain frequencies of the visible spectrum, which constitute its <u>emission spectrum</u>.

If the same element, also in gaseous state, receives electromagnetic radiation, it absorbs in certain frequencies of the visible spectrum, <u>precisely the same ones in which it emits when it is stimulated by heat</u>. This will be its <u>absorption spectrum.</u>

<h2>This means that every element absorbs radiation in the same wavelengths in which it emits it. </h2>

You might be interested in

I’m confused on this worksheet can someone help

Finger [1]

Answer:

well for C the horse is at rest at 10 to 15

D is 0 to 10

E is 15 to 25

Explanation:

If you search up examples it will help with this worksheet.

If I am work then I appologize

5 0

3 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$