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Dual Nature of Radiation and Matter

The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Given, cut-off voltage, V_o = 1.5 V

Maximum kinetic energy of photoelectrons emitted is,

K . E ._{\max} = {eV}_{0} = 1.5 \times 1.6 \times 10^{- 19} J = 2.4 \times 10^{- 19} J

1520 Views

The work function of caesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴Hz is incident on the metal surface, photoemission of electrons occurs. What is the stopping potential?

Work function of caesium metal = 2.14 eV
frequency of light wave = 6 x 10¹⁴Hz

The stopping potential can be calculated as,

{eV}_{0} = K . E ._{\max}

\Rightarrow

V_{0} = \frac{K . E ._{\max}}{e} = \frac{0.35 eV}{e} = 0.35 V

2711 Views

The work function of caesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴Hz is incident on the metal surface, photoemission of electrons occurs. What is the maximum speed of the emitted photoelectrons?

Given, a ceasium metal.
Work function = 2.14 eV
Frequency of light,

ν

= 6 x 10¹⁴Hz

Maximum kinetic energy is given by,

\frac{1}{2} {mv}^{2}_{\max} = 0.346 eV = 0.346 \times 1.6 \times 10^{- 19} J

\Rightarrow

v_{\max} = \sqrt{\frac{0.346 \times 1.6 \times 10^{- 19} \times 2}{9.1 \times 10^{- 31}}} {ms}^{- 1}

= 3.488 \times 10^{5} {ms}^{- 1} = 348.8 km s^{- 1}

i.e.,

v_{\max} = 349 km s^{- 1} .

which is the required maximum speed of the photoelectrons.

1425 Views

The work function of caesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴Hz is incident on the metal surface, photoemission of electrons occurs. What is the maximum kinetic energy of the emitted electrons?

Given,
Work function of caesium metal = 2.14 eV
Frequency of light,

ν

= 6 x 10¹⁴Hz

Maximum kinetic energy of the emitted electrons,

K . E ._{\max} = hv - ϕ_{0} = \frac{6.63 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} - 2.14

E_{\max} = 0.346 eV = 0.35 eV

1616 Views

Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.

Here,
Voltage, V = 30 kV = 30 x 10³V = 3 x 10⁴V

(a) Using the formula,

{hv}_{\max} = eV

v_{\max} = \frac{eV}{h} = \frac{1.6 \times 10^{- 19} \times 3 \times 10^{4}}{6.63 \times 10^{- 34}} Hz

= 7.24 \times 10^{18} Hz

is the required maximum frequency.

(b) Minimum wavelength of x-rays produced,

λ_{m i n} = \frac{c}{v_{\max}} = \frac{3 \times 10^{8}}{7.24 \times 10^{18}} m

\Rightarrow

λ_{m i n} = 0.414 \times 10^{- 10} m = 0.414 Å

= 0.0414 nm .

1977 Views

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Previous Year Papers

Dual Nature of Radiation and Matter

Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.

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Previous Year Papers

Dual Nature of Radiation and Matter

Find the(a) maximum frequency, and(b) minimum wavelength of X-rays produced by 30 kV electrons.

Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.