Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Dual Nature of Radiation question

2016 · Shift 1 · Q43
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Physics
  4. /Dual Nature of Radiation
  5. /2016 · Shift 1 · Q43

Dual Nature of Radiation question

2016 · Shift 1 · Q43

JEE AdvancedPhysicsDual Nature of RadiationMCQ+3 / −1
In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength (λ\lambdaλ) of incident light and the corresponding stopping potential (V0) are given below:

λ(μm)\lambda \left( {\mu m} \right)λ(μm) V0(Volt)
0.3 2.0
0.4 1.0
0.5 0.4


Given that c = 3 ×\times× 108 ms-1 and e = 1.6 ×\times× 10-19 C, Planck's constant (in units of J-s) found from such an experiment is) :
  1. A
    6.0 ×\times× 10-34
  2. B
    6.6 ×\times× 10-34
  3. C
    6.4 ×\times× 10-34
  4. D
    6.8 ×\times× 10-34
View written solutionFree

Correct answer: C

  1. Use Einstein’s photoelectric equation

For photoelectric emission, eV0=hν−ϕ=hcλ−ϕeV_0 = h\nu - \phi = \frac{hc}{\lambda} - \phieV0​=hν−ϕ=λhc​−ϕ

So, V0=hce⋅1λ−ϕeV_0 = \frac{hc}{e}\cdot \frac{1}{\lambda} - \frac{\phi}{e}V0​=ehc​⋅λ1​−eϕ​

This is a straight-line relation between V0V_0V0​ and 1λ\dfrac{1}{\lambda}λ1​ with slope m=hcem = \frac{hc}{e}m=ehc​

Hence, h=mech = \frac{me}{c}h=cme​

  1. Calculate the slope from the data

Given:

  • λ1=0.3 μm=0.3×10−6 m\lambda_1 = 0.3\,\mu m = 0.3 \times 10^{-6}\,mλ1​=0.3μm=0.3×10−6m, V01=2.0 VV_{01}=2.0\,VV01​=2.0V
  • λ2=0.4 μm=0.4×10−6 m\lambda_2 = 0.4\,\mu m = 0.4 \times 10^{-6}\,mλ2​=0.4μm=0.4×10−6m, V02=1.0 VV_{02}=1.0\,VV02​=1.0V
  • λ3=0.5 μm=0.5×10−6 m\lambda_3 = 0.5\,\mu m = 0.5 \times 10^{-6}\,mλ3​=0.5μm=0.5×10−6m, V03=0.4 VV_{03}=0.4\,VV03​=0.4V

Take two points, say (0.3 μm,2.0V)(0.3\,\mu m, 2.0V)(0.3μm,2.0V) and (0.4 μm,1.0V)(0.4\,\mu m, 1.0V)(0.4μm,1.0V).

First compute 1λ\dfrac{1}{\lambda}λ1​: 1λ1=10.3×10−6=3.33×106 m−1\frac{1}{\lambda_1} = \frac{1}{0.3\times 10^{-6}} = 3.33\times 10^6\,m^{-1}λ1​1​=0.3×10−61​=3.33×106m−1 1λ2=10.4×10−6=2.5×106 m−1\frac{1}{\lambda_2} = \frac{1}{0.4\times 10^{-6}} = 2.5\times 10^6\,m^{-1}λ2​1​=0.4×10−61​=2.5×106m−1

So slope is m=ΔV0Δ(1/λ)=2.0−1.0(3.33−2.5)×106m = \frac{\Delta V_0}{\Delta (1/\lambda)} = \frac{2.0-1.0}{(3.33-2.5)\times 10^6}m=Δ(1/λ)ΔV0​​=(3.33−2.5)×1062.0−1.0​ m=1.00.83×106≈1.2×10−6 V⋅mm = \frac{1.0}{0.83\times 10^6} \approx 1.2\times 10^{-6}\,V\cdot mm=0.83×1061.0​≈1.2×10−6V⋅m

Check with another pair:

Between 0.4 μm0.4\,\mu m0.4μm and 0.5 μm0.5\,\mu m0.5μm, 1λ3=10.5×10−6=2.0×106 m−1\frac{1}{\lambda_3} = \frac{1}{0.5\times 10^{-6}} = 2.0\times 10^6\,m^{-1}λ3​1​=0.5×10−61​=2.0×106m−1 m=1.0−0.4(2.5−2.0)×106=0.60.5×106=1.2×10−6 V⋅mm = \frac{1.0-0.4}{(2.5-2.0)\times 10^6} = \frac{0.6}{0.5\times 10^6} = 1.2\times 10^{-6}\,V\cdot mm=(2.5−2.0)×1061.0−0.4​=0.5×1060.6​=1.2×10−6V⋅m

So the slope is consistent: m=1.2×10−6 V⋅mm = 1.2\times 10^{-6}\,V\cdot mm=1.2×10−6V⋅m

  1. Now compute Planck’s constant

Using h=mech = \frac{me}{c}h=cme​

Substitute: h=(1.2×10−6)(1.6×10−19)3×108h = \frac{(1.2\times 10^{-6})(1.6\times 10^{-19})}{3\times 10^8}h=3×108(1.2×10−6)(1.6×10−19)​

h=1.92×10−253×108h = \frac{1.92\times 10^{-25}}{3\times 10^8}h=3×1081.92×10−25​

h=0.64×10−33=6.4×10−34 J⋅sh = 0.64\times 10^{-33} = 6.4\times 10^{-34}\,J\cdot sh=0.64×10−33=6.4×10−34J⋅s

  1. Match with the options

h=6.4×10−34 J⋅sh = 6.4\times 10^{-34}\,J\cdot sh=6.4×10−34J⋅s

Therefore, the correct option is C.

  1. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So the derived answer agrees with the stored correct answer.

PreviousNext

More from Dual Nature of Radiation

  • Light of wavelength λ ph falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is ϕ and the anode is a wire mesh of conducting material kept at a distance d from the… Includes diagram2016 · Multiple correct
  • For photo-electric effect with incident photon wavelength λ, the stopping potential is V0. Identify the correct variation(s) of V0 with λ and λ1​.2015 · Multiple correct
  • An electron in an excited state of Li2+ ion has angular momentum 2π3h​. The de Broglie wavelength of the electron in this state is p π a0 (where a0 is the Bohr radius). The value of p is2015 · Numerical
  • A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1 : u2 = 2 : 1 and hc = 1240 eV…2014 · MCQ
  • The work functions of silver and sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is ​.2013 · Numerical
  • Using the expression 2dsinθ=λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90o. The wavelength λ is exactly known and the error in θ is constant for all…2013 · Multiple correct
  • A proton is fired from very far away towards a nucleus with charge Q = 120e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is ​…2012 · Numerical
  • The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is 109 s. The mass of an atom of this radioisotope is 10 − 25 kg. The mass (in mg) of the radioactive sample is ​…2011 · Numerical