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uysha [10]
2 years ago
8

A laser shines violet light with a frequency of 703 thz onto the metal with a threshold frequency of 295 thz, what is the kineti

c energy of the ejected electrons in ev? chegg
Physics
1 answer:
hram777 [196]2 years ago
3 0

The kinetic energy of the ejected electrons will be 2782.5 ×10⁻²² eV.

<h3>What is threshold frequency?</h3>

The threshold frequency of incoming radiation is the lowest frequency at which photoelectric emission or electron emission is impossible.

The threshold frequency is the light frequency that causes an electron to dislodge and emit from the metal's surface.

From the photoelectric effect, the equation obtained as;

\rm  hv=hv_0+KE\\\\ KE=h(v-v_0)\\\\ KE=6.625 \times 10^{-34}(702-282)Hz \\\\ KE= 6.625 \times 10^{-34}\times 420 \times 10^{12} \ Hz \\\\ KE=2782 .5 \times 10^{-22} \ eV

Hence,the kinetic energy of the ejected electrons will be 2782.5 ×10⁻²² eV.

To learn more about the threshold frequency, refer to the link;

brainly.com/question/2499414

#SPJ1

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But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

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this means, for the angular momentum measure from the rod:

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| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)

but the angle is 90°, so:

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|

| \vec{L}_{rod_i} | = r_i * m * v_i

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}

| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}

For our final angular momentum we have:

| \vec{L}_{rod_f} | = r_f * m * v_f

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| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f

but, as the angular momentum is constant, this must be equal to the initial angular momentum

7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f

v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}

v_f = 15 \frac{m}{s}

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