Question

2.2. Ross 128 is a nearby M-dwarf star. Ross 128 has a mass of only 0.17 M◉ and a luminosity of 0.0036 L◉. It has a photospheric temperature of ~3,190 K.

(a) What is its wavelength of peak emission in µm? And in nanometers (nm)?

(b) What part of the electromagnetic spectrum is this?

(c) What color would you expect Ross 128 to appear to the naked eye?

(d) Which star of the two stars (i.e., R136a1 or Ross 128) do you think makes for a more promising host star for habitable planets? State two reasons.

Answer 1

(A)according to wiens displacement law

Peak Wavelenght lamda max=2900/T

Where T is temperature...through this formula we will get wavelenght in micrometer

T=3190 K given so

Lamda max=2900/3190 micrometer

Lamda max=0.91micrometer

1 micrometer=1000 nm

So 0.91 micrometer=0.91×1000nm=910nm

(B)700 to 10000nm is the range of infrared em spectrum.so 910nm will be infrared spectrum.

(C)Infrared are generally invisible to human naked eye but wavelenght upto 1050nm can be seen in and it is red in color.

(D)R136a1 will be a more promising host star beacause

(A)its luminosity 8710000L is much higher than Ross128 luminosity0.0036L,means r136a1 can provide more energy than ross 128.

(B)r136a1 surface temperature is 53000 K where ross 128 is 3190k...the more the temperature of a star it will be more stable.