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aliina [53]
3 years ago
15

I need help with this question ​

Chemistry
1 answer:
VARVARA [1.3K]3 years ago
6 0

Answer:

2.25×10¯³ mm.

Explanation:

From the question given above, we obtained the following information:

Diameter in micrometer = 2.25 μm

Diameter in millimetre (mm) =?

Next we shall convert 2.25 μm to metre (m). This can be obtained as follow:

1 μm = 1×10¯⁶ m

Therefore,

2.25 μm = 2.25 μm / 1 μm × 1×10¯⁶ m

2.25 μm = 2.25×10¯⁶ m

Finally, we shall convert 2.25×10¯⁶ m to millimetre (mm) as follow:

1 m = 1000 mm

Therefore,

2.25×10¯⁶ m = 2.25×10¯⁶ m /1 m × 1000 mm

2.25×10¯⁶ m = 2.25×10¯³ mm

Therefore, 2.25 μm is equivalent to 2.25×10¯³ mm.

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b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

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e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

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Frequency = speed of light/wavelength

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E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

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Ecalculated × Avogadros constant

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As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

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e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

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