(C) According to the photoelectric effect equation: $h\nu = KE_{max} + h\nu_0$ Where $h$ is Planck's constant $(6.63 \times 10^{-34} \ J \cdot s)$,$\n

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(C) According to the photoelectric effect equation: $h\nu = KE_{max} + h\nu_0$ Where $h$ is Planck's constant $(6.63 \times 10^{-34} \ J \cdot s)$,$\nu$ is the incident frequency,$KE_{max}$ is the maximum kinetic energy,and $\nu_0$ is the threshold frequency. Rearranging for $\nu_0$: $h\nu_0 = h\nu - KE_{max}$ $\nu_0 = \nu - \frac{KE_{max}}{h}$ Substituting the given values: $\nu_0 = (4 \times 10^{14} \ s^{-1}) - \frac{6.63 \times 10^{-20} \ J}{6.63 \times 10^{-34} \ J \cdot s}$ $\nu_0 = 4 \times 10^{14} \ s^{-1} - 1 \times 10^{14} \ s^{-1}$ $\nu_0 = 3 \times 10^{14} \ s^{-1}$

Class 11 Chemistry Structure of Atom Atomic models and Planck's quantum theory

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If the maximum kinetic energy of photoelectrons ejected from a metal surface when it is irradiated with a radiation of frequency $4 \times 10^{14} \ s^{-1}$ is $6.63 \times 10^{-20} \ J$,then the threshold frequency of the metal is

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A
$2 \times 10^{14} \ s^{-1}$
B
$1 \times 10^{14} \ s^{-1}$
C
$3 \times 10^{14} \ s^{-1}$
D
$1 \times 10^{-14} \ s^{-1}$

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Structure of Atom

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Atomic models and Planck's quantum theory

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MediumJEE MAIN 2014

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If the maximum kinetic energy of photoelectrons ejected from a metal surface when it is irradiated with a radiation of frequency $4 \times 10^{14} \ s^{-1}$ is $6.63 \times 10^{-20} \ J$,then the threshold frequency of the metal is

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If the work function for the photoelectron emission of a metal is $3.75 \ eV$,then the threshold wavelength of the radiation needed for the ejection of the electron is approximately (in $nm$)

For ionising an excited hydrogen atom,the required energy in $eV$ will be -

Consider the following spectral lines for atomic hydrogen:$A$. First line of Paschen series$B$. Second line of Balmer series$C$. Third line of Paschen series$D$. Fourth line of Bracket seriesThe correct arrangement of the above lines in ascending order of energy is:

An excited hydrogen atom emits light in the ultraviolet region at a frequency of $2.47 \times 10^{15} \ Hz$. Calculate the energy of a single photon. (Given: $h = 6.63 \times 10^{-34} \ J \cdot s$)

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Consider the following spectral lines for atomic hydrogen:
$A$. First line of Paschen series
$B$. Second line of Balmer series
$C$. Third line of Paschen series
$D$. Fourth line of Bracket series
The correct arrangement of the above lines in ascending order of energy is: