Question

Need quality answer for the above question..

→ Explaination required
→ Don't post spam answers​

Answer 1

EXPLANATION.

Electron ejected from the surface of a metal.

Light of certain frequency is incident on it, are stopped fully by a retarding potential of = 3 volts.

Frequency = 6 x 10¹⁴ S⁻¹.

As we know that,

⇒ h = 6 x 10⁻³⁴ j-sec.

⇒ Charge on electron = 1.6 x 10⁻¹⁹C.

Stopping potential = (hυ - hο)/e.

⇒ e x stopping potential = h(υ - υο).

⇒ 1.6 x 10⁻¹⁹ x 3 = 6 x 10⁻³⁴(υ - 6 x 10¹⁴).

⇒ 4.8 x 10⁻¹⁹ = (6 x 10⁻³⁴)υ - (6 x 10¹⁴ x 6 x 10⁻³⁴).

⇒ 4.8 x 10⁻¹⁹ = (6 x 10⁻³⁴)υ - (36 x 10⁻²⁰).

⇒ 4.8 x 10⁻¹⁹ + 36 x 10⁻²⁰ = (6 x 10⁻³⁴)υ.

⇒ 10⁻¹⁹(4.8 + 36 x 10⁻¹) = (6 x 10⁻³⁴)υ.

⇒ 10⁻¹⁹(4.8 + 3.6) = (6 x 10⁻³⁴)υ.

⇒ 10⁻¹⁹(8.4) = (6 x 10⁻³⁴)υ.

⇒ υ = (8.4 x 10⁻¹⁹)/(6 x 10⁻³⁴).

⇒ υ = 1.4 x 10⁻¹⁹ x 10³⁴.

⇒ υ = 1.4 x 10¹⁵.

⇒ υ = 14 x 10¹⁴.

Option [C] is correct answer.

Answer 2

• $\boxed{\sf{\purple{Option\:\bold{3)\:14\:\times\:10^{14}\:\sf{is\:correct!}}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬

Explanation :

$\underline{\underline{\bf{\green{Given\::-}}}}$

• $\small\sf Frequency_{(photo\:electric\:effect)}\:=\:\bf{6\:\times\:10^{14}s^{-1}}$
• $\small\sf Charge\: on\: the\: electron = \bf{1.6\:\times\:10^{-19}C}$
• $\small\sf Retarding \:potential = \bf{3\:volts}$
• $\small\sf h = \bf{6\:\times\:10^{-34}\:J-sec}$

$\underline{\underline{\bf{\green{To\:Find\::-}}}}$

• $\sf Frequency_{(incident \:light)}\: in \:s^{-1}\:=\:?$

$\underline{\underline{\bf{\green{Solution\::-}}}}$

$\small\longrightarrow\:\sf Stopping \:potential = (h\nu - h{\nu}_{0})/e$

$\small\longrightarrow\:\sf Stopping \:potential \:\times\:e = h(\nu - {\nu}_{0})$

$\small\longrightarrow\:\sf 3 \:\times\:1.6\:\times\:10^{-19} = 6\:\times\:10^{-34} (\nu - 6\:\times\:10^{14})$

$\small\longrightarrow\:\sf 4.8\times\:10^{-19} = (6\:\times\:10^{-34})\nu - (6\:\times\:10^{14}\:\times\:6\:\times\:10^{-34})$

$\small\longrightarrow\:\sf 4.8\times\:10^{-19} = (6\:\times\:10^{-34})\nu - (6\:\times\:6\:\times\:10^{14}\:\times\:10^{-34})$

$\small\longrightarrow\:\sf 4.8\times\:10^{-19} = (6\:\times\:10^{-34})\nu - (36\:\times\:10^{(-34 + 14)})$

$\small\longrightarrow\:\sf 4.8\times\:10^{-19} = (6\:\times\:10^{-34})\nu - (36\:\times\:10^{-20})$

$\small\longrightarrow\:\sf (4.8\times\:10^{-19}) - (36\:\times\:10^{-20}) = (6\:\times\:10^{-34})\nu$

$\small\longrightarrow\:\sf 10^{-19}(4.8 + 36\:\times\:10^{-1}) = (6\:\times\:10^{-34})\nu$

$\small\longrightarrow\:\sf 10^{-19}(4.8 + 3.6) = (6\:\times\:10^{-34})\nu$

$\small\longrightarrow\:\sf 10^{-19}(8.4) = (6\:\times\:10^{-34})\nu$

$\small\longrightarrow\:\sf 8.4 \:\times\:10^{-19} = (6\:\times\:10^{-34})\nu$

$\small\longrightarrow\:\sf \dfrac{8.4 \:\times\:10^{-19}}{6\:\times\:10^{-34}} = \nu$

$\small\longrightarrow\:\sf \dfrac{\cancel{8.4}\:\times\:10^{-19}}{\cancel{6}\:\times\:10^{-34}} = \nu$

$\small\longrightarrow\:\sf \dfrac{1.4 \:\times\:10^{-19}}{1\:\times\:10^{-34}} = \nu$

$\small\longrightarrow\:\sf \dfrac{1.4 \:\times\:10^{-19}}{10^{-34}} = \nu$

$\small\longrightarrow\:\sf 1.4 \:\times\:10^{-19}\:\times\:10^{34} = \nu$

$\small\longrightarrow\:\sf 1.4 \:\times\:10^{(-19 + 34)} = \nu$

$\small\longrightarrow\:\sf 1.4 \:\times\:10^{15} = \nu$

$\small\longrightarrow\:\sf \nu = 14 \:\times\:10^{14}$

$\small\therefore\:\underline{\underline{\bf{\red{Frequency_{(incident \:light)}\: in \:s^{-1} = 14 \:\times\:10^{14}}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬