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प्रश्न
The figure shows a plot of stopping potential (V0) versus `1/lambda`, where λ is the wavelength of the radiation causing photoelectric emission from a surface. The slope of the line is equal to ______.
विकल्प
Φ0
`h/e`
`(hc)/e`
`(h^2c)/e^2`
उत्तर
The figure shows a plot of stopping potential (V0) versus `1/lambda`, where λ is the wavelength of the radiation causing photoelectric emission from a surface. The slope of the line is equal to `underlinebb(h/e)`.
Explanation:
`1/lambda = ν` (frequency)
The photoelectric effect in terms of stopping potential is as follows:
hν = hν0 + eV
Writing this equation in terms of y = mx + c, we get
V = `((hν - hν_0))/e`
On comparing this equation with y = mx + c, we get
`m = h/e`
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(a) Identify the variable X on the horizontal axis.
(b) What does the point A on the horizontal axis represent?
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(d) Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I1 > I2 > I3) having same frequency.
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Answer the following question.
Plot a graph of photocurrent versus anode potential for radiation of frequency ν and intensities I1 and I2 (I1 < I2).
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The photoelectric current increases with increase of intensity of incident light.
The graph shows the variation of photocurrent for a photosensitive metal
- What does X and A on the horizontal axis represent?
- Draw this graph for three different values of frequencies of incident radiation ʋ1, ʋ2 and ʋ3 (ʋ3 > ʋ2 > ʋ1) for the same intensity.
- Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I3 > I2 > I1) having the same frequency.
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