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frosja888 [35]
1 year ago
6

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

Physics
2 answers:
OlgaM077 [116]1 year ago
6 0

Answer:

1.05 × 10 ∧15 Hz.

Explanation:

The minimum amount of energy of photon E = 6.94×10∧-19  J

The value of Plank's constant , h = 6.626 × 10∧-34 J s

So the minimum frequency of light necessary to emit electrons from titanium via photoelectric effect , E = h . ν

where ν is the 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.

erma4kov [3.2K]1 year ago
4 0

<span>E = h x f </span>

<span>. . . then : </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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\dfrac{P_1V_b}{T_1}=\frac{P_2V_s}{T_2}\\\Rightarrow \dfrac{V_b}{V_s}=\frac{P_2T_1}{P_1T_2}\\\Rightarrow \dfrac{V_b}{V_s}=\frac{1\times 277.15}{3.5\times 296.15}\\\Rightarrow \dfrac{V_s}{V_b}=0.26738^{-1}\\\Rightarrow \dfrac{V_s}{V_b}=3.73994

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