What is the temperature of a star whose maximum light is emitted at a wavelength of 290 nm?

Physics

1 answer:

viva [34]3 years ago

7 0

Answer:

Temperature, T = 10000 K

Explanation:

It is given that,

Wavelength, $\lambda=290\ nm=290\times 10^{-9}\ m$

We need to find the temperature of a star whose maximum light is emitted at this wavelength. It can be calculated using Wein's displacement law as :

$\lambda_{max}.T=2.9\times 10^{-3}$

$T=\dfrac{2.9\times 10^{-3}}{290\times 10^{-9}}$

T = 10000 K

So, the temperature of a star is 10000 K. Hence, this is the required solution.

You might be interested in

Which type of elements are likely to have no electical charge at all

boyakko [2]

Answer:

Noble gas element

Explanation:

because Noble gas element can neither loose nor gain electrons

6 0

4 years ago

Read 2 more answers

Hello, how to do qn 5?:)l

Hitman42 [59]

Answer:

Explanation:

Please see the attached picture for the full solution.

Since we are only concerned about the decrease in gravitational potential energy of the car, we look at the decrease in height of the car as it moves from point X to point Y, instead of the distance travelled by the car.

7 0

3 years ago

A particle has a charge of -4.25 nC.

SpyIntel [72]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C$

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

$E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m$