Solutions for Exercises of Dual Nature of Radiation and Matter
1) What will be the energy of each photon in monochromatic light of frequency 5 x1014 Hz?
???? = 5 ×1014Hz, E = ?
Energy of each photon is given as,
E = hν
∴ E = 6.6 × 10-34 × 5 × 1014
∴ E = 3.3 × 10-19 J
2) Observations from an experiment on photoelectric effect for the stopping potential by varying the incident frequency were plotted. The slope of the linear curve was found to be approximately 4.1 x10-15 V s. Given that the charge of electron is 1.6 x10-19, find the value of Plank’s constant h.
We know that for the photoelectric experiment,
hν = W0 + K.E.max
Here, ν is the frequency of incident light, W0 is the work function and K.E.max is the maximum kinetic energy of emitted photoelectron.
We also know that K.E.max is balanced by the stopping potential
i.e. K.E.max=eV0
∴ hν = W0 + eV0
$latex \therefore\;V_0=\left(\frac he\right)v-\left(\frac{W_0}e\right)$
Hence if we draw the curve between stopping potential V and incident frequency ν then we get a straight line with y intercept (W0/e) and slope (h/e).
But according to the question slope of the linear curve was found to be approximately 4.1 x10-15 Vs.
$latex \therefore\;\left(\frac he\right)\;=\;4.1\;\times\;10^{-15}$
$latex \therefore\;h\;=\;4.1\;\times\;10^{-15}\;\times\;1.6\;\times\;10^{-19}$
$latex i.e.\;h\;=\;6.56\;\times\;10^{-3}\;Js$
3) The threshold wavelength of tungsten is 2.76 x 10-5cm. (a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 x 10-5 cm. (b) What will be the maximum kinetic energy of electrons ejected in each of the following cases (i) if ultraviolet radiation of wavelength λ = 1.80 × 10-5 cm and (ii) radiation of frequency 4×1015Hz is made incident on the tungsten surface.
λ = 2.76× 10-5cm
i.e λ= 2.76 × 10-7m,
a) λ > 2.76× 10-5cm
For photoelectric effect to happen, incident frequency of the light must be greater than the threshold frequency
i.e. ν > ν0 but v = c/λ hence,
$latex \frac c\lambda>\frac c{\lambda_0}\;\;\;\;\;\;\therefore\lambda<\lambda_0$
Hence incident wavelength must be lesser than the threshold wavelength for the photoelectric effect. But in the question incident wavelength is greater than the threshold hence no photoelectric effect.
b) We have, K.E.max = hν − W0
$latex K.E._{max}=\frac{hc}\lambda-hv_0$
$latex =\frac{hc}\lambda-\frac{hc}{\lambda_0}$
$latex \therefore\;K.E.{}_{max}=hc\left[\frac1\lambda-\frac1{\lambda_0}\right]$
i) λ = 1.80 × 10-5cm = 1.80 × 10-7m
∴ K.E.max,
$latex =\;6.6\times10^{-34}\times3\times10^8\left[\frac1{1.80\times10^{-7}}-\frac1{2.76\times10^{-7}}\right]$
$latex \therefore\;K.E._{max}\;=\;19.8\;\times\;10^{-26}\;\times\;\frac1{10^{-7}}\left[\frac1{1.80}-\frac1{2.76}\right]$
$latex K.E._{max}\;=\;3.82\;\times\;10^{-1}J$
$latex K.E._{max}\;=\;\frac{3.82\times10^{-19}}{1.6\times10^{-1}}eV$
$latex K.E._{max}\;=\;2.40\;eV$
ii) ν = 4 × 1015Hz
$latex K.E._{max}\;=\;hv-\frac{hc}{\lambda_0}$
$latex K.E._{max}\;=\;h\left(v-\frac c{\lambda_0}\right)$
$latex K.E._{max}\;=\;6.6\times10^{-34}\left[4\times10^{15}-\frac{3\times10^8}{2.76\times10^{-7}}\right]$
$latex K.E._{max}\;=\;6.6\times10^{-34}\left[4\times10^{15}-1.08\times10^{15}\right]\\$
∴ K.E.max = 1.98 × 10-18J = 12.17eV
4) Photocurrent recorded in the micro ammeter in an experimental set-up of photoelectric effect vanishes when the retarding potential is more than 0.8 V if the wavelength of incident radiation is 4950 Å. If the source of incident radiation is changed, the stopping potential turns out to be 1.2 V. Find the work function of the cathode material and the wavelength of the second source.
V01 = 0.8V, λ1 = 4950Ȧ = 4950×10-10m
V02 = 1.2V, λ1 =?, W0=?
For the photoelectric effect we have,
$latex \;K.E_{max}\;=\;hv\;-\;W_0\\$
$latex \therefore eV_0\;=\;\frac{hc}\lambda-W_0$
For first wavelength,
$latex eV_{01}=\frac{hc}{\lambda_1}-W_0\;……(1)$
For second wavelength,
$latex eV_{02}=\frac{hc}{\lambda_2}-W_0\;……(2)$
Work function will be same as material is same.
Subtracting equation 1 from equation 2, we have,
$latex e\left(V_{02}-V_{01}\right)=hc\left[\frac1{\lambda_2}-\frac1{\lambda_1}\right]$
$latex \therefore\;\left[\frac1{\lambda_2}-\frac1{\lambda_1}\right]=\frac{e\left(V_{02}-V_{01}\right)}{hc}$
$latex =\;\frac{1.6\times10^{-19}\left(1.2-0.8\right)}{19.8\times10^{-26}}$
$latex \therefore\;\left[\frac1{\lambda_2}-\frac1{\lambda_1}\right]=3.23\times10^5$
$latex \therefore\frac1{\lambda_2}=\left(3.23\times10^5\right)+\frac1{4950\times10^{10}}$
$latex \therefore\frac1{\lambda_2}=\left(3.23\times10^5\right)+\left(20.2\times10^5\right)$
$latex \frac1{\lambda_2}=\;23.43\;\times\;10^5$
$latex \therefore\;\lambda_2=\frac1{23.43\times10^5}$
$latex \therefore\;\lambda_2\;=\;4.270\;\times\;10^{-7}m$
$latex \therefore\;\lambda_2\;=\;=\;4270\;\overset\circ A\;$
Put this value of λ in equation (1) we have,
$latex 1.6\times10^{-19}\times0.8=\frac{19.8\times10^{-26}}{4.270\times10^{-7}}-W_0$
$latex \therefore\;W_0\;=\;(4.637\times\;10^{-19})-\left(1.28\times10^{-19}\right)$
$latex \therefore\;W_0\;=\;3.357\;\times\;10^{-19}J\;=\;2eV$
5) Radiation of wavelength 4500 Å is incident on a metal having work function 2.0 eV. Due to the presence of a magnetic field B, The most energetic photoelectrons emitted in a direction perpendicular to the field move along a circular path of radius 20 cm. What is the value of the magnetic field B?
$latex \lambda=4500\;\overset\circ A=4500\times10^{-10}m,$
B = ?
For photoelectric effect we have,
K.E.max = hv – W0
$latex \therefore\;K.E._{max}=\frac{hc}\lambda-W_0$
$latex =\frac{19.8\times10^{-26}}{4500\times10^{-1}}-3.2\times10^{-19}$
$latex =\;(4.4\;\times\;10^{-19})\;-\;(3.2\;\times\;10^{-19})$
$latex \therefore\;K.E._{max}=1.2\;\times\;10^{-19}J$
Now the emitted photoelectron with this kinetic energy is entered into magnetic field. Hence it will perform circular motion with radius given as,
$latex r\;=\;\frac{\sqrt{2m\left(K.E.\right)}}{eB}$
$latex i.e.\;B\;=\;\frac{\sqrt{2m\left(K.E.\right)}}{er}$
$latex \therefore\;B\;=\;\frac{2\times9.1\times10^{-31}\times1.2\times10^{-19}}{1.6\times10^{-19}\times20\times10^{-2}}$
$latex \therefore\;B\;=\;\frac{4.673\times10^{-25}}{32\times10^{-21}}$
$latex \therefore\;B\;=\;1.47\;\times\;10^{-5}T$
6) Given the following data for incident wavelength and the stopping potential obtained from an experiment on photoelectric effect, estimate the value of Planck’s constant and the work function of the cathode material. What is the threshold frequency and corresponding wavelength? What is the most likely metal used for emitter?
V01 = 1.95V, λ1 = 2536Ȧ = 2536 × 10-10m,
V02 = 0.5V, λ2 = 3650Ȧ = 3650 × 10-10m
h =?,W0 =?ν0 =?,λ0 =?
For the photoelectric effect we have,
$latex K.E._{max}\;=\;hv\;-\;W_0$
$latex \therefore\;eV_0\;=\;\frac{hc}\lambda\;-\;W_0$
For first wavelength,
$latex \therefore\;eV_{01}\;=\;\frac{hc}{\lambda_1}\;-\;W_0\;……(1)$
For second wavelength,
$latex \therefore\;eV_{02}\;=\;\frac{hc}{\lambda_2}\;-\;W_0\;……(2)$
Work function will be same as material is same.
Subtracting equation 2 from equation 1, we have,
$latex e\left(V_{02}\;-\;V_{01}\right)\;=\;hc\left[\frac1{\lambda_1}-\frac1{\lambda_2}\right]$
$latex \therefore\;1.6\;\times\;10^{-19}(1.95\;-\;0.5)$
$latex =\;h\;\times\;3\;\times\;10^8\;\times\;\left[\frac1{2536\times10^{-10}}-\frac1{3650\times10^{-1}}\right]$
$latex \therefore\;2.32\;\times\;10^{-19}=\frac{h\times3\times10^8}{10^{-10}}\times\;\left[\frac1{2536}-\frac1{3650}\right]$
$latex \therefore\;2.32\;\times\;10^{-19}\;=\;3\;\times\;10^{18}\;\times\;h\;\times\;1.203\;\times\;10^{-4}$
$latex \therefore\;h\;=\frac{2.32\times10^{-19}}{3\times10^{18}\times1.203\times10^{-4}}$
$latex \therefore\;h\;=\;6.42\;\times\;10^{-34}\;Js$
Put value of h in equation (1), we have,
$latex 1.6\times10^{-19}\times1.95=\frac{6.42\times10^{-34}\times3\times10^8}{2536\times10^{-10}}-W_0$
∴ W0 = 4.474 × 10-19J = 2.8eV
We have, $latex W_0\;=\;h\nu_0\;\;\;\;\;\;\;\;\;\therefore v_0=\frac{W_0}h$
$latex \therefore\;\nu_0\;=\;\frac{4.474\times10^{-1}}{6.42\times10^{-34}}$
$latex \therefore\;\nu_0\;=\;6.86\;\times\;10^{14}Hz$
$latex \therefore\;\lambda_0\;=\;\frac c{v_0}$
$latex \therefore\;\lambda_0\;=\;\frac{3\times10^8}{6.86\times10^{14}}$
$latex \therefore\;\lambda_0\;=\;4.415\;\times\;10^{-7}m$
$latex \therefore\;\lambda_0\;=\;=\;4415\;\overset\circ A$
Best material for the emitter is calcium since it has reasonably low work function.
7) Calculate the wavelength associated with an electron, its momentum and speed (a) when it is accelerated through a potential of 54 V, (b) when it is moving with kinetic energy of 150 eV.
a) V = 54 V
For an electron accelerated by potential V
we have,
$latex \therefore\;\lambda\;=\;\frac{12.28\;\times\;10^{-10}}{\sqrt V}m$
$latex \therefore\;\lambda\;=\;\frac{12.28\;\times\;10^{-10}}{\sqrt{54}}m$
$latex \therefore\;\lambda\;=\;1.671\;\times\;10^{-10}m\;=\;0.167nm$
$latex \;\lambda\;=\;\frac hp\;\;\;\;\;\;\;\;\therefore p\;=\;h\lambda$
$latex \therefore\;p\;=\frac{6.6\times10^{-34}}{1.671\times10^{-10}}$
$latex \therefore\;p\;=3.949\;\times\;10^{-24}\;kgms^{-1}$
$latex p\;=\;mv\;\;\;\;\;\;\;\;\therefore\;v\;=\;\frac pm$
$latex \therefore\;v\;=\;\frac{3.949\times10^{-24}}{9.1\times10^{-31}}$
$latex \therefore\;v\;=\;4.363\;\times\;10^6m/s$
b) K.E.= EK = 150eV = 240 × 10-19J
For an electron moving with kinetic energy EK
$latex \lambda\;=\;\frac h{\sqrt{2mE_K}}$
$latex \lambda\;=\;\frac{6.6\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times240\times10^{-19}}}$
$latex \lambda\;=\;\frac{6.6\times10^{-34}}{\sqrt{4368\times10^{-50}}}$
$latex \lambda\;=\;\frac{6.6\times10^{-34}}{\sqrt{66.1\times10^{-25}}}$
$latex \lambda\;=\;0.1\;\times\;10^{-9}m\;=\;0.1nm$
$latex \lambda\;=\;\frac hp\;\;\;\;\;\;\;\;\therefore p\;=\;\frac h\lambda$
$latex \therefore\;p\;=\;\frac{6.6\times10^{-34}}{0.1\times10^{-9}}$
$latex \therefore\;p\;=\;6.613\;\times\;10^{-24}kgms^{-1}$
$latex p\;=\;mv\;\;\;\;\;\;\;\;\therefore v\;=\;\frac pm$
$latex \therefore v\;=\;\frac{6.613\times10^{-24}}{9.1\times10^{-31}}$
$latex \therefore v\;=\;7.26\;\times\;10^6m/s$
8) The de Broglie wavelengths associated with an electron and a proton are same. What will be the ratio of (i) their momenta (ii) their kinetic energies ?
$latex \lambda_e=\lambda_p,\;\frac{p_e}{p_p}=?,\;\frac{E_e}{E_p}=?$
We know,
De Broglie wavelength is given as,$latex \lambda\;=\;\frac hp$
$latex \lambda_e\;=\;\frac h{p_e}$ and $latex \lambda_p\;=\;\frac h{p_p}$ But, $latex \lambda_e\;=\;\lambda_p$
$latex \therefore\;\frac h{p_e}=\frac h{p_p}\;\;\;\;\;\;\;\;\therefore\frac{p_e}{p_p}=1$
Also,$latex p=\sqrt{2mE}\;…here\;E\;is\;K.E.$
$latex \therefore\;p\;=\;\sqrt{2m_eE_e}\;\&$
$latex \therefore\;p_p\;=\;\sqrt{2m_pE_p}\;$
$latex \therefore\frac{\sqrt{2m_pE_p}\;}{\sqrt{2m_pE}}\;=\;1$
$latex \therefore\frac{m_eE_e\;}{m_pE_e}\;=\;1$
$latex \therefore m_eE_e\;=\;m_pE_p$
$latex \therefore\;\frac{E_p}{E_p}=\frac{m_p}{m_e}$
$latex \therefore\;\frac{E_e}{E_p}\;=\;\frac{1.67\times10^{-27}}{9.1\times10^{-31}}$
$latex \therefore\;\frac{E_e}{E_p}\;=\;1836$
9) Two particles have the same de Broglie wavelength and one is moving four times as fast as the other. If the slower particle is an α-particle, what are the possibilities for the other particle?
λ1 = λ2, v = 4v2,
We know, De Broglie wavelength is given as,
$latex \lambda\;=\;\frac hp$
and as λ1 = λ2 we have p1 = p2
∴ m1v1 = m2v2
∴ m14v2 = m2v2
$latex i.e.\;m_1\;=\;\frac{m_2}4$
Given that second particle is alpha particle. And we can see that first particle is having one fourth times. Hence first particle must be a neutron or a proton as they both have mass one fourth of the mass of alpha particle.
10) What is the speed of a proton having de Broglie wavelength of 0.08 Å?
λ = 0.08Å = 0.08 × 10-10m,
We know, De Broglie wavelength is given as,
$latex \lambda\;=\;\frac hp\;\;\;\;\;\;\;\;\therefore\;p\;=\frac h\lambda$
$latex i.e.\;p\;=\frac{6.6\;\times\;10^{-34}}{0.08\times10^{-10}}$
$latex i.e.\;p\;=82.5\;\times\;10^{-24}kgms^{-1}$
But, p = mv
i.e v = p/m
$latex \therefore\;v\;=\;\frac{82.5\times10^{-34}}{1.67\times10^{-27}}$
$latex \therefore\;v\;=\;49.519\;\times\;10^3m/s$
11) In nuclear reactors, neutrons travel with energies of 5 x 10-21 J. Find their speed and wavelength.
K.E.= E = 5 × 10-21J, v = ? , λ = ?
$latex K.E.=E=\frac12mv^2$
$latex \therefore\;v=\sqrt{\frac{2E}m}$
$latex \therefore\;v=\sqrt{\frac{2\times5\times10^{-21}}{1.67\times10^{-27}}}$
$latex \therefore\;v=\sqrt{5.988\;\times\;10^6}$
∴ v = 2.45 × 103m/s
For an particle moving with kinetic energy E,
$latex \lambda\;=\;\frac h{\sqrt{2mE}}$
$latex \therefore\;\lambda\;=\frac{6.6\times10^{-34}}{\sqrt{2\times1.67\times10^{-27}\times5\times10^{-21}}}$
$latex \therefore\;\lambda\;=\frac{6.6\times10^{-34}}{4.086\times10^{-24}}$
∴ λ = 1.625 × 10-10m
i.e. λ = 1.625Ȧ
12) Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the (a) same speed, (b) same energy and (c) same momentum? State which of the two will have the longer wavelength in each case?
a) ve = vp
Momenta of electron and proton are given as,
pe = meve and pp = mpvp
$latex \therefore\;\frac{p_e}{p_p}=\frac{m_ev_e}{m_pv_p}=\frac{m_e}{m_p}$
$latex =\;\frac{9.1\;\times\;10^{-31}}{1.67\;\times\;10^{-27}}$
$latex =\;5.449\;\times\;10^{-4}$
We know, De Broglie wavelength is given as,
$latex \lambda\;=\;\frac hp$
$latex \therefore\;\frac{\lambda_e\;}{\lambda_p}=\;\frac{p_p}{p_e}$
$latex =\;\frac1{5.449\;\times10^{-4}}$
$latex \therefore\;\frac{\lambda_e\;}{\lambda_p}=\;\frac{p_p}{p_e}\;=\;1836$
Electron has longer wavelength.
b)Ee= Ep
For a particle moving with kinetic energy E, De Broglie wavelength is given as,
$latex \lambda\;=\;\frac h{\sqrt{2mE}}$
$latex as\;\lambda\;\propto\;\frac1{\sqrt m}\;as\;E\;is\;cons\tan t\;$
$latex \therefore\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{1.67\times10^{-27}}{9.1\times10^{-31}}}$
$latex i.e.\;\frac{\lambda_e}{\lambda_p}=\sqrt{1836}$
$latex \therefore\;\frac{\lambda_e}{\lambda_p}=\;42.85$
Electron has longer wavelength.
c) pe = pp
We know, De Broglie wavelength is given as,
$latex \lambda\;=\;\frac hp$
$latex \therefore\;\frac{\lambda_e}{\lambda_p}\;=\;\frac{p_p}{p_e}\;=\;1$
Electron and proton have same wavelength.