A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to (a) H (b) H^1/2 (c) H⁰ (d) H^-1/2

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Pallavi Pathak

Contributor-Level 10

Explanation- velocity of a freely falling body is v= $\sqrt[]{2 g h}$

And $? = \frac{h}{m v} = \frac{h}{m \sqrt[]{2 g h}}$

$? = h$ ^-1

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New answer posted

8 months ago

NEET Dual Nature of Radiation and Matter

Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.
(i) Estimate number of photons emitted by the bulb per second. [Assume no other losses]
(ii) Will there be photoelectric emission?

(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was

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Pallavi Pathak

Contributor-Level 10

Explanation-number of photon emitted per second n= $\frac{p}{\frac{h c}{?}} = \frac{p ?}{h c} = \frac{20 * 5000 * 10^{- 10}}{6.62 * 10^{- 34} * 3 * 10^{8}}$

$= 5 * 10^{19} s^{- 1}$

(ii) E=hc $/ ?$ = $\frac{6.62 * 10^{- 34} * 3 * 10^{8}}{5000 * 10^{- 10} * 1.6 * 10^{- 19}} = 2.48 e V$ this enegy is greater than 2 so emission is possible

(iii) work function $?$ = $\frac{p}{4 ? d^{2}} * ? r^{2} ? t$ = $?_{o}$

$? t$ = $\frac{4 ? d^{2}}{p r^{2}}$ = $\frac{4 * 2 * 16 * 1.6 * 10^{- 19} * 2^{2}}{20 * {(1.5 * 10^{- 10})}^{- 2}} = 28.4 s$

(iv) N= $\frac{n ? r^{2}}{4 ? d^{2}} * ? t$

= $\frac{5 * 10^{19} * {(1.5 * 10^{- 10})}^{2} * 28.4}{4 * ({2)}^{2}}$ =2

(v) as the time of emission is 11.04s so photoelectric is not spontaneous.

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