Question

radiation Lambda equals to 2800 angstrom is incident on an isolated metallic sphere of radius 4.5 sentimeter work function of the metal is 2.4 electron volt the maximum charge on the metal surfaces​

Answer 1

maximum charge on the metal surface is 2.7 × 10^-16 C

from photoelectric effect,

K.E = hc/λ - Φ

where h is Plank's constant, c is speed of light , λ is wavelength and Φ is work function.

given, Φ = 2.4 eV

λ = 2800A°

using shortcut formula, hc/λ = 12400/λ(in A°) eV

so, K.E = 12400/2800 - 2.4

= 2.028 eV

now, 1eV = 1.6 × 10^-19 J

so, K.E = 2.028 × 1.6 × 10^-19 J

= 3.2448 × 10^-19 J

we know, qV = K.E

here V is potential.

for isolated metallic sphere , V = kq/r

so, q × (kq/r) = K.E

⇒kq²/r = 3.2448 × 10^-19 J

⇒9 × 10^9 × q²/(4.5 × 10^-2) = 3.2448 × 10^-19

⇒q² = 3.2448 × 10^-19 × (4.5 × 10^-2)²/9 × 10^9

⇒q = 2.7 × 10^-16 C

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Answer 2

$\huge\bold\purple{Answer:-}$

from photoelectric effect,

K.E = hc/λ - Φ

where h is Plank's constant, c is speed of light , λ is wavelength and Φ is work function.

given, Φ = 2.4 eV

λ = 2800A°

using shortcut formula, hc/λ = 12400/λ(in A°) eV

so, K.E = 12400/2800 - 2.4

= 2.028 eV

now, 1eV = 1.6 × 10^-19 J

so, K.E = 2.028 × 1.6 × 10^-19 J

= 3.2448 × 10^-19 J

we know, qV = K.E

here V is potential.

for isolated metallic sphere , V = kq/r

so, q × (kq/r) = K.E

⇒kq²/r = 3.2448 × 10^-19 J

⇒9 × 10^9 × q²/(4.5 × 10^-2) = 3.2448 × 10^-19

⇒q² = 3.2448 × 10^-19 × (4.5 × 10^-2)²/9 × 10^9

⇒q = 2.7 × 10^-16 C