JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 18)
The stopping potential V0 (in volt) as a function of frequency ($$\upsilon $$) 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)_12th_April_Morning_Slot_en_18_1.png)
(Given: Planck’s constant (h) = 6.63 × 10–34 Js, electron charges e = 1.6 × 10–19 C)
1.95 eV
2.12 eV
1.82 eV
1.66 eV
Explanation
$$hv = \varphi + e{v_0}$$
$${v_0} = {{hv} \over e} - {\varphi \over e}$$
v0 is zero for v = 4 × 1014 Hz
$$0 = {{hv} \over e} - {\varphi \over e}$$
$$ \Rightarrow \phi = hv$$
$$ = {{6.63 \times {{10}^{ - 34}} \times 4 \times {{10}^{14}}} \over {1.6 \times {{10}^{ - 19}}}} = 1.66\,eV$$
$${v_0} = {{hv} \over e} - {\varphi \over e}$$
v0 is zero for v = 4 × 1014 Hz
$$0 = {{hv} \over e} - {\varphi \over e}$$
$$ \Rightarrow \phi = hv$$
$$ = {{6.63 \times {{10}^{ - 34}} \times 4 \times {{10}^{14}}} \over {1.6 \times {{10}^{ - 19}}}} = 1.66\,eV$$
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