Question 173: Dual Nature of Matter and Radiation

Question

The stopping potential V
0
(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)

Accepted Answer

$$hv = \varphi  + e{v_0}$$
$${v_0} = {{hv} \over e} - {\varphi  \over e}$$
v
0
is zero for v = 4 ÃÂ 10
14
Hz
$$0 = {{hv} \over e} - {\varphi  \over e}$$
$$ \Rightarrow \phi  = hv$$
$$ = {{6.63 	imes {{10}^{ - 34}} 	imes 4 	imes {{10}^{14}}} \over {1.6 	imes {{10}^{ - 19}}}} = 1.66\,eV$$

Answer

1.95 eV

Answer

2.12 eV

Answer

1.82 eV

Answer

1.66 eV

Dual Nature of Matter and Radiation

The stopping potential V 0 (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)

Choose the correct answer: