## 1A.3

$c=\lambda v$

Martina
Posts: 89
Joined: Thu Jul 25, 2019 12:16 am

### 1A.3

Which of the following happens when the frequency of electromagnetic radiation decreases? Explain your reasoning.
(a) The speed of the radiation decreases.
(b) The wavelength of the radiation decreases.
(c) The extent of the change in the electrical field at a given point
decreases.
(d) The energy of the radiation increases.

Can someone please explain the logic behind the answer to this question?

lilymayek_1E
Posts: 84
Joined: Sat Aug 17, 2019 12:16 am

### Re: 1A.3

(a) from v(wave)=c, it must be true that the speed of light (electromagnetic radiation) is constant. thus, it's impossible for speed to decline
(b) this is also false because of v(wave)=c. Because c is a constant, frequency and wavelength must have an inverse relationship to maintain this constant. so if frequency decreases, the wavelength must increase, contrary to the statement
(c) this is true. think of an electrical field like a wave (a sine wave, for example). when frequency decreases (this would be visually seen as the waves on a graph "stretching out"), in order to maintain the same speed, the wavelength must increase (stretch). the peaks and troughs of the waves would be closer to the "x-axis" on a graph (on a graph, the slope would decrease), which signifies that change in the field (from this "zero value, or the "x-axis") has declined.
*sorry if that's a bit confusing, you almost need to draw out a graph in order to explain it!*
(d) no; from the equation E=hv, because h is a constant (cannot change), decreasing frequency would decrease the overall energy.

hope this helps somewhat!!!

Hannah Lee 2F
Posts: 99
Joined: Thu Jul 11, 2019 12:15 am

### Re: 1A.3

(a) The speed of the radiation decreases.
Speed of radiation is constant and is described as speed of light, where c = 3.00 x 10^8 m/s.

(b) The wavelength of the radiation decreases.
Wavelength would actually increase, because frequency and wavelength share an inversely proportional relationship, as described by c = λv. When v decreases, λ will increase, and vice versa.

(c) The extent of the change in the electrical field at a given point decreases.
In other words, "extent of the change in the electrical field" is the slope of the wave. Since wavelengths increase and get wider as frequency decreases, the slope of the wave would also decrease, as would the energy of the photon. This makes C correct.

(d) The energy of the radiation increases.
No, because the energy of radiation increases linearly with frequency according to E = hv, where E = energy per photon. If frequency decreased, then the energy would also decrease.

lauraxie2e
Posts: 87
Joined: Fri Aug 09, 2019 12:17 am

### Re: 1A.3

I believe the answer is C because E=hv and so frequency and energy are directly related.

Amy Xiao 1I
Posts: 85
Joined: Sat Jul 20, 2019 12:15 am

### Re: 1A.3

C would be the correct answer. Think of the "change in the electrical field" as slope-- how steep a wave goes. As frequency decreases, the waves stretch out farther and the slope would be less steep because they are now farther apart.