Question

5. What is the work function (w.) of the metal whose threshold frequency (Vo) is 5.2 x 1014-15
(1) 3.44 * 10
(2) 3.44 × 10-19 J
(3) 1.44 × 10-17)
(4) 1.44 % 1017 J
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Answer 1

Answer : The work function is, $3.44\times 10^{-19}J$

Explanation : Given,

Threshold frequency = $5.2\times 10^{14}s^{-1}$

Formula used :

$E=h\nu_o$

where,

E = work function of metal

h = Planck's constant = $6.626\times 10^{-34}Js$

$\nu_o$ = threshold frequency

Now put all the given values in this formula, we get the value of work function.

$E=(6.626\times 10^{-34}Js)\times (5.2\times 10^{14}s^{-1})$

$E=3.44\times 10^{-19}J$

Therefore, the work function is, $3.44\times 10^{-19}J$

Answer 2

Answer:

Explanation:

What is the work function (wo) of the metal whose threshold frequency (v) is 5.2 x 1014 5-1?

solution : Work function of a metal is the amount of energy required to remove an electron from metal surface.

according to Plank's theory,

work function = plank's constant × threshold frequency

given, Threshold frequency = 5.2 × 10¹⁴ Hz

we know, plank's constant = 6.63 × 10^-34 J.s

so, work function = 6.63 × 10^-34 × 5.2 × 10¹⁴ = 34.476 × 10^-20 J

= 3.4476 × 10^-19 J

hence, work function orThreshold energy of the metal is 3.4476 × 10^-19 J