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QJEEM19Phd8c8: JEE Main 2019 Physics - Class 12 JEEM19Phd8c8 Solution: The stopping potential V 0 (in volt) as a function of frequency …

The stopping potential V0 (in volt) as a function of frequency ( υ\upsilon ) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be:
(Given: Planck’s constant (h) = 6.63 × 10–34 Js, electron charges e = 1.6 × 10–19 C) JEE Main 2019 (Online) 12th April Morning Slot Physics - Dual Nature of Radiation Question 139 English

Multiple Choice Options

  • A 1.95 eV
  • B 2.12 eV
  • C 1.82 eV
  • D 1.66 eV

Step-by-Step Verified Solution

hv=φ+ev0hv = \varphi + e{v_0}

v0=hve−φe{v_0} = {{hv} \over e} - {\varphi \over e}

v0 is zero for v = 4 × 1014 Hz

0=hve−φe0 = {{hv} \over e} - {\varphi \over e}

⇒ϕ=hv\Rightarrow \phi = hv

=6.63×10−34×4×10141.6×10−19=1.66 eV= {{6.63 \times {{10}^{ - 34}} \times 4 \times {{10}^{14}}} \over {1.6 \times {{10}^{ - 19}}}} = 1.66\,eV

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