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Atoms and Nuclei question

2014 · Shift 0 · Q48
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Atoms and Nuclei question

2014 · Shift 0 · Q48

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
The radiation corresponding to 3→23 \to 23→2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field 3×10−4 T.3 \times {10^{ - 4}}\,T.3×10−4T. If the radius of the larger circular path followed by these electrons is 10.0mm10.0mm10.0mm, the work function of the metal is close to:
  1. A
    1.8eV1.8eV1.8eV
  2. B
    1.1eV1.1eV1.1eV
  3. C
    0.8eV0.8eV0.8eV
  4. D
    1.6eV1.6eV1.6eV
View written solutionFree

Correct answer: B

  1. Energy of the incident photon

The radiation is due to the hydrogen transition 3→23 \to 23→2. For hydrogen, En=−13.6n2 eVE_n=-\frac{13.6}{n^2}\,\text{eV}En​=−n213.6​eV

So, E3=−13.69,E2=−13.64E_3=-\frac{13.6}{9}, \qquad E_2=-\frac{13.6}{4}E3​=−913.6​,E2​=−413.6​

Hence photon energy is hν=E3−E2=13.6(14−19)h\nu = E_3-E_2 = 13.6\left(\frac{1}{4}-\frac{1}{9}\right)hν=E3​−E2​=13.6(41​−91​) hν=13.6(536)=1.89 eVh\nu = 13.6\left(\frac{5}{36}\right)=1.89\,\text{eV}hν=13.6(365​)=1.89eV

  1. Kinetic energy of emitted photoelectrons from magnetic radius

The electron enters magnetic field B=3×10−4 TB=3\times 10^{-4}\,\text{T}B=3×10−4T and follows circular path of radius r=10.0 mm=1.0×10−2 mr=10.0\,\text{mm}=1.0\times 10^{-2}\,\text{m}r=10.0mm=1.0×10−2m

For motion perpendicular to magnetic field, r=mveBr=\frac{mv}{eB}r=eBmv​ so v=eBrmv=\frac{eBr}{m}v=meBr​

Kinetic energy, K=12mv2=12m(eBrm)2=e2B2r22mK=\frac{1}{2}mv^2=\frac{1}{2}m\left(\frac{eBr}{m}\right)^2=\frac{e^2B^2r^2}{2m}K=21​mv2=21​m(meBr​)2=2me2B2r2​

Substitute values: e=1.6×10−19 C,m=9.1×10−31 kge=1.6\times 10^{-19}\,\text{C},\quad m=9.1\times 10^{-31}\,\text{kg}e=1.6×10−19C,m=9.1×10−31kg

K=(1.6×10−19)2(3×10−4)2(1.0×10−2)22(9.1×10−31)K=\frac{(1.6\times 10^{-19})^2(3\times 10^{-4})^2(1.0\times 10^{-2})^2}{2(9.1\times 10^{-31})}K=2(9.1×10−31)(1.6×10−19)2(3×10−4)2(1.0×10−2)2​

Now, (1.6×10−19)2=2.56×10−38(1.6\times 10^{-19})^2=2.56\times 10^{-38}(1.6×10−19)2=2.56×10−38 (3×10−4)2=9×10−8(3\times 10^{-4})^2=9\times 10^{-8}(3×10−4)2=9×10−8 (1.0×10−2)2=1×10−4(1.0\times 10^{-2})^2=1\times 10^{-4}(1.0×10−2)2=1×10−4

Thus, K=2.56×9×10−5018.2×10−31K=\frac{2.56\times 9\times 10^{-50}}{18.2\times 10^{-31}}K=18.2×10−312.56×9×10−50​ K≈23.04×10−5018.2×10−31K\approx \frac{23.04\times 10^{-50}}{18.2\times 10^{-31}}K≈18.2×10−3123.04×10−50​ K≈1.27×10−19 JK\approx 1.27\times 10^{-19}\,\text{J}K≈1.27×10−19J

Convert into eV: K=1.27×10−191.6×10−19≈0.79 eVK=\frac{1.27\times 10^{-19}}{1.6\times 10^{-19}}\approx 0.79\,\text{eV}K=1.6×10−191.27×10−19​≈0.79eV

  1. Apply photoelectric equation

hν=ϕ+Kmax⁡h\nu = \phi + K_{\max}hν=ϕ+Kmax​

So work function, ϕ=hν−Kmax⁡\phi = h\nu - K_{\max}ϕ=hν−Kmax​ ϕ=1.89−0.79=1.10 eV\phi = 1.89 - 0.79 = 1.10\,\text{eV}ϕ=1.89−0.79=1.10eV

  1. Match with options

Closest option is: 1.1 eV\boxed{1.1\,\text{eV}}1.1eV​ So the correct option is B.

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