Determine the MA based on the diagram below.

A photon has momentum of magnitude 8.30×10−28 kg⋅m/s . Part APart complete What is the energy of this photon? Give your answer i

d1i1m1o1n [39]

Answer with Explanation:

We are given that

Momentum of photon= $8.3\times 10^{-28} kg.m/s$

a. We have to find the energy of this photon.

Speed of photon= $c=3\times 10^8 m/s$

We know that

Momentum=p= $\frac{h}{\lambda}$

Where

h= $6.63\times 10^{-34}$ J-s=Plank's constant

$\lambda=$ Wavelength of photon

$\lambda=\frac{h}{p}$

$\lambda=\frac{6.63\times 10^{-34}}{8.3\times 10^{-28}}$

$\lambda=7.99\times 10^{-7}$ m

$E=\frac{hc}{\lambda}$

$E=\frac{6.63\times 10^{-34}\times 3\times 10^8}{7.99\times 10^{-7}}$

$E=2.49\times 10^{-19}$ J

Hence, the energy of photon= $2.49\times 10^{-19}$ J

B.Energy of photon in electron volt= $\frac{2.49\times 10^{-19}}{1.6\times 10^{-19}}=1.55 eV$

Energy of photon= $1.55eV$

C.Wavelength of photon = $\lambda=7.99\times 10^{-7}m$

6 0

4 years ago

What evidence suggests that the ancestors of whales once walked on land?

aliina [53]

C whales have similar DNA to elephants

3 0

2 years ago

Read 2 more answers

A marathon runner runs at a steady 15 km/hr. When the runner is 7.5 km from the finish, a bird begins flying from the runner to

Whitepunk [10]

Answer:

The value is $D = 15 \ km$

Explanation:

From the question we are told that

The speed of the marathon runner is $v = 15 \ km /hr$

The distance from the distance from the finish is $d = 7.5 \ km$

The speed of the bird is $v_b = 30 \ km / hr$

Generally the time taken for the runner to reach the finish is mathematically represented as

$t = \frac{d}{v}$

$t = \frac{7.5}{15}$

$t = \frac{1}{2}$

So the distance covered by the bird is

$D = v_b * t$

$D = 30 * \frac{1}{2}$

$D = 15 \ km$

6 0

3 years ago

A protostar's radius decreases by a factor of 100 and its surface temperature increases by a factor of two before it becomes a m

Sliva [168]

Answer:

$L_f = K (\frac{r}{100})^2 * (2T)^4$

$L_f = K \frac{r^2}{10000} * 16 T^4$

$L_f = \frac{16}{10000} k r^2 T^4 = \frac{1}{625} k r^2 T^4$

$L_f = \frac{1}{625} L_i$

So then we see that the final luminosity decrease by a factor of 625 so then the correct answer for this case would be:

B. Decreases by a factor of 625

Explanation:

For this case we can use the formula of luminosity in terms of the radius and the temperature given by:

$L_i = K r^2 T^4$

Where L_i = initial luminosity, r= radius and T = temperature.

We know that we decrease the radius by a factor of 100 and the temperature increases by a factor of 2 so then the new luminosity would be:

$L_f = K (\frac{r}{100})^2 * (2T)^4$

$L_f = K \frac{r^2}{10000} * 16 T^4$

$L_f = \frac{16}{10000} k r^2 T^4 = \frac{1}{625} k r^2 T^4$

$L_f = \frac{1}{625} L_i$

So then we see that the final luminosity decrease by a factor of 625 so then the correct answer for this case would be:

B. Decreases by a factor of 625

6 0

3 years ago

What is the half-life of an isotope if after 30 days you have 31.25 g remaining from a 250 g beginning sample size?

Dima020 [189]

Answer:The time required for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life, T1/2, and the decay constant is given by T1/2 = 0.693/λ.

Explanation:

4 0

2 years ago