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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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So, we have:

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So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F

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2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

q_1' = q_1\\q_2' = q_2\\d' = 2d

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3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

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q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d

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F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F

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The mechanical energy can also be transformed into electrical energy through a sort of dynamo system in vehicles. Stereo players use the electrical energy to produce sound.

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