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kenny6666
3 months ago
6

What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect?

Physics
2 answers:
ValentinkaMS [3.4K]3 months ago
6 0

Answer:

1.05 × 10^15 Hz.

Explanation:

The minimal energy of the photon is E = 6.94×10^-19  J.

Planck's constant is noted as h = 6.626 × 10^-34 J s.

Thus, the minimum frequency necessary to emit electrons from titanium through the photoelectric effect is given by E = h. ν.

Where ν represents frequency.

                                     ν = E / h

                                     ν = 6.94×10^-19  J / 6.626 × 10^-34 J s

                                         = 1.05 × 10^15 / s

                                         = 1.05 × 10^15 Hz.

kicyunya [3.2K]3 months ago
4 0

<span>E = h x f </span>

<span>Thus: </span>

<span>f = E / h </span>
<span>f = 4.41•10^-19 / 6.62•10^-34 </span>
<span>f = 6.66•10^14 Hz (s^-1) </span>


<span>b/ What is the wavelength of this light? </span>
<span>------------------------------ </span>

<span>λ = c / f </span>
<span>λ = 3•10^8 / 6.66•10^14 </span>
<span>λ = 4.50•10^-7 m </span>
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