The maximum kinetic energy ( E_k ) of emitted photoelectrons against frequency \nu of incident radiation is plotted as shown in fig. The slope of the graph is equal to

Answer is C. E_{k}=h\nu -\phi So, comparing it with y=mx+c we get slope =h

Dual Nature Of Radiation And Matter

Q: Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).Electrons accelerated by potential V are diffracted from a crystal. If d =1A^o and i = 30^{\circ} , V should be about : (h=6.6 \times10^{-34}js, m=9.1\times10^{-31}kg, e=1.6\times10^{-19}c)

We have, 2 d sin \theta = n \lambda angle of incidence = 90 - i Given i = 30^o , d = 1 \times 10^{-10} m , h = 6.6 \times 10^{-34} , m = 9.1 \times 10^{-31} and e = 1.6 \times 10^{-19} Also n = 1 for 1^{st} order diffraction.Therefore, 2 \times 1 \times 10^{-10} sin(90-30) = 1 \times \lambda \implies \lambda = 2 \times 10^{-10} \dfrac {\sqrt {3}}{2} = \sqrt {3} \times 10^ {-10}m We have de-Broglie wavelength, \lambda = \dfrac {h}{mv} and also \dfrac {1}{2} mv^2 = eV \implies v = \sqrt{ \dfrac{2eV}{m}} Substituting in de-Broglie wavength, we get, \lambda = \dfrac {h} { \sqrt {2meV} } = \sqrt {3} \times 10^{-10} m \implies V = 49.86 V

Q: With the decrease in the wave length of the incident radiation the velocity of the photoelectrons emitted from a given metal

As wavelength decrease , frequency increases and hence K.E of photoelectrons increases , thereby increasing the velocity.

Q: The frequency of a photon associated with an energy of 3.31 eV is (given h =6.625\times 10^{-34} Js)

energy = h\nu 3.31eV = 6.625*10^{-34}\nu \nu = \dfrac{3.31*1.6*10^{-19}}{6.625*10^{-34}} ( \because e =1.6*10^{-19}C ) = 0.8*10^{15}Hz

Q: An X-Ray tube operates at a p.d. of 12400\ V . The wavelength of the highest frequency X-Rays emitted by this tube is nearly:( hc=1240\ eV-nm )

Energy=p.d\times e = 12400V\times e =12400\ eV Energy=\dfrac{hc}{\lambda } \lambda =\dfrac{hc}{energy} =\dfrac{1240}{12400}\ nm =0.1\ nm =1 \ A^{0}

Q: The graph between the de Broglie wavelength and the momentum of a photon is a

The relationship between de Broglie wavelength (\lambda) and the momentum (p) of a photon is given by: \lambda=\dfrac{h}{p} . So, the graph between the de Broglie wavelength and the momentum of a photon is a rectangular hyperbola.

Q: The given graph shows the variation of \sqrt{f} vs Z for characteristics X-rays. Lines 1, 2, 3, 4 shown in the graph corresponds to any one of K_{\alpha}, K_{\beta}, L_{\alpha}, L_{\beta}, then L_{\beta} is represented by (f = frequency, Z = Atomic number )

For a given value of Z, energy of K-series is more than L-series, also \beta −line has more energy as compared to \alpha −line of L-series.Hence, L_{\beta} will be represented by line 3.

Q: A photoelectric cell is connected to a source of variable potential difference, connected across it and the photoelectric current resulting (\mu A) is plotted against the applied potential difference (V). The graph in the broken line represents one for a given frequency and intensity of the incident radiation. If the frequency is increased and the intensity is reduced, then which of the following graphs of unbroken line represents the new situation?

Since the intensity is reduced to half, means the number of photon falling per unit area becomes half. So the photoelectric current will decreases .So D is correct option. But by increasing frequency stopping Potential will shift more into the negative x-axis .

Q: In a photoelectric emission, electrons are ejected from metals X and Y by light of frequency f. The potential difference V required to stop the electrons is measured for various frequencies. If y has a greater work function than X, which graph illustrates the expected results?

The answer is option A. First, the gradient of the graph cannot change (always = h/e ), so answers are A or D.If Y has a greater work function than X, then the graph for Y should have a more negative y-intercept. therefore figure in option A depicts the concept.

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Problem solving tips

4 min read

Dual Nature Of Radiation And Matter

- Want to score better marks in exams? Have a look at some problem solving tips.

Tip-1: Calculating the kinetic energy of photoelectrons
When light of suitable energy is incident on a metal surface, photoelectrons are emitted from the surface. many times we are asked to calculate the maximum kinetic energy of the photoelectrons.
Maximum Kinetic energy of photoelectrons = Energy of the incident light - Work function

Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).
Electrons accelerated by potential V are diffracted from a crystal. If

d = 1 A^{o}

and

i = 3 0^{\circ}

, V should be about :

(h = 6.6 \times 1 0^{- 34} j s, m = 9.1 \times 1 0^{- 31} k g, e = 1.6 \times 1 0^{- 19} c)

2000

50

500

1000

View solution

With the decrease in the wave length of the incident radiation the velocity of the photoelectrons emitted from a given metal

Remains same

Increases

Decreases

Increases first and then decreases

View solution

Tip-2: Converting energy into the unit of electron-volt
We usy denote energy in the unit of joule. But in case of photoelectric emission or X-ray production, it is convenient to use energy in terms of electron-volt. We get energy in electron volt by dividing the energy in joule by the charge of electron.

1 e V = 1.6 \times 1 0^{- 19} volt

.uall

The frequency of a photon associated with an energy of 3.31 eV is (given h

= 6.625 \times 1 0^{- 34}

Js)

0.8 \times 1 0^{15}

1.6 \times 1 0^{15}

3.2 \times 1 0^{15}

8.0 \times 1 0^{15}

View solution

Tip-3: Determining hard X-rays and soft X-rays
Hard X-rays have high energy. One can easily determine the energy of X-ray by measuring its wavelength and vice-versa. The maximum possible energy of X-ray = Kinetic energy of electrons moving in a potential. Finally,